CHEM 105 - ELEMENTS OF

GENERAL CHEMISTRY -

Wave-particle duality and de Broglie

wavelength

Question Bank - Set 3

Liberty University

Question 1

Question

An electron is accelerated through a potential difference of 100 V. Calculate

the de Broglie wavelength associated with this electron. (Take the charge of an

electron to be −1.6×10−19 C and the mass of an electron to be 9.11 ×10−31

kg.)

Solution

1. We can use the energy of the electron to find its velocity, and then use

this velocity to calculate its de Broglie wavelength.

2. The energy (E) of the electron can be calculated using the potential dif-

ference (V) through the equation:

E=eV

where eis the charge of the electron.

3. Substituting the given values, we get:

E= (−1.6×10−19 C) ×(100 V) = −1.6×10−17 J

4. The kinetic energy of the electron can be equated to its total energy:

KE =E=1

2mv2

where mis the mass of the electron and vis its velocity.

5. Rearranging the equation gives:

v=r2E

m=s2× −1.6×10−17 J

9.11 ×10−31 kg

6. Calculating the velocity gives:

v≈5.92 ×106m/s

7. The de Broglie wavelength (λ) can be calculated using the equation:

λ=h

mv

where his the Planck constant.

8. Substituting the values of h,m, and v, we get:

λ=6.63 ×10−34 J s

(9.11 ×10−31 kg) ×(5.92 ×106m/s)

9. Calculating the de Broglie wavelength gives:

λ≈1.22 ×10−10 m

Question 2

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron. (Hint: The energy of the electron can be

related to its de Broglie wavelength through the formula E=p2

2m, where Eis

the kinetic energy, pis the momentum, and mis the mass of the electron.)

Solution

Step 1: We know that the kinetic energy of the electron is given by E=qV ,

where qis the charge of the electron and Vis the potential difference. Since the

charge of an electron is 1.6×10−19 C and the potential difference is 100 V, the

kinetic energy of the electron is

E= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 2: The kinetic energy of the electron can also be expressed in terms

of momentum as E=p2

2m, where pis the momentum and mis the mass of the

electron. Therefore, we have

p2

2m= 1.6×10−17 J

2

p2= 3.2×10−17 ×m

Step 3: The de Broglie wavelength of the electron is given by λ=h

p, where

his Planck’s constant. Substituting p=√3.2×10−17 ×minto the de Broglie

wavelength formula, we get

λ=h

√3.2×10−17 ×m

Step 4: The mass of an electron is 9.11 ×10−31 kg. Substituting this value

and h= 6.63 ×10−34 J s into the de Broglie wavelength formula, we find

λ=6.63 ×10−34

√3.2×10−17 ×9.11 ×10−31

λ≈7.27 ×10−11 m

Therefore, the de Broglie wavelength of the electron accelerated through a

potential difference of 100 V is approximately 7.27 ×10−11 m.

Question 3

Question

An electron with a kinetic energy of 200 eV is moving in vacuum. Find the de

Broglie wavelength associated with this electron.

Solution

Step 1: Convert the kinetic energy from electron-volts to joules.

1 eV = 1.6×10−19 J

So, the kinetic energy of the electron in joules is:

E= 200 eV ×1.6×10−19 J/eV = 3.2×10−17 J

Step 2: Use the de Broglie wavelength formula to find the wavelength asso-

ciated with the electron. The de Broglie wavelength is given by:

λ=h

p

where λ= wavelength, h= Planck’s constant (6.626 ×10−34 J s), p= momen-

tum = √2mE,m= mass of the electron (9.11 ×10−31 kg), E= kinetic energy

of the electron.

Step 3: Calculate the momentum of the electron.

p=p2×9.11 ×10−31 kg ×3.2×10−17 J

3

p=q5.82 ×10−16 kg m2s−2

p= 7.63 ×10−8kg m/s

Step 4: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 J s

7.63 ×10−8kg m/s

λ≈8.68 ×10−11 m

Therefore, the de Broglie wavelength associated with the electron with a

kinetic energy of 200 eV is approximately 8.68 ×10−11 meters.

Question 4

Question

A beam of electrons with kinetic energy 1 keV is incident on a crystal lattice.

Assuming that the electrons behave as de Broglie waves, calculate the de Broglie

wavelength of the electrons.

Solution

Step 1: Calculate the momentum of the electrons. The momentum of an

electron can be calculated using the formula:

p=√2mE

where mis the mass of the electron and Eis the kinetic energy.

Given:

E= 1 keV = 1.6×10−16 J

m= 9.11 ×10−31 kg

Substitute the given values into the formula:

p=p2×9.11 ×10−31 ×1.6×10−16

p=p2×9.11 ×1.6×10−47

p=p2×14.576 ×10−47

p=p29.152 ×10−47

p= 5.401 ×10−24 kg m/s

Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength of

an electron is given by:

λ=h

p

4

where his the Planck constant.

The Planck constant is h= 6.626 ×10−34 m2kg/s.

Substitute the values of hand pinto the formula:

λ=6.626 ×10−34

5.401 ×10−24

λ=6.626

5.401 ×10−10

λ= 1.227 ×10−10 m

Therefore, the de Broglie wavelength of the electrons is 1.227 ×10−10 m.

Question 5

Question

An electron is accelerated through a potential difference of 120 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference

provided. Step 2: Use the kinetic energy to find the velocity of the electron. Step

3: Apply the de Broglie wavelength formula to find the wavelength associated

with the electron.

Step 1: Calculate the kinetic energy of the electron. The kinetic energy

(KE) of an electron accelerated through a potential difference ∆Vis given by:

KE =e·∆V

where eis the elementary charge (1.6×10−19 C) and ∆Vis the potential

difference (120 V).

KE = (1.6×10−19 C) ×(120 V) = 1.92 ×10−17 J

Step 2: Calculate the velocity of the electron. The kinetic energy of the

electron can be expressed in terms of its velocity vas:

KE =1

2mv2

where mis the mass of the electron (9.11 ×10−31 kg). Solving for v:

v=r2KE

m=r2×1.92 ×10−17

9.11 ×10−31 ≈4.08 ×106m/s

5

Step 3: Calculate the de Broglie wavelength. The de Broglie wavelength

(λ) of a particle is given by:

λ=h

p

where his the Planck constant (6.63 ×10−34 J·s) and pis the momentum of

the electron. The momentum of the electron is given by p=mv, where mis

the mass of the electron and vis the velocity.

p= (9.11 ×10−31 kg) ×(4.08 ×106m/s) ≈3.72 ×10−24 kg ·m/s

λ=6.63 ×10−34

3.72 ×10−24 ≈1.78 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 120 V is approximately 1.78 ×10−10 m.

Question 6

Question

An electron is accelerated through a potential difference of 100 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

The potential energy gained by the electron is given by qV , where qis the charge

of the electron (1.6 ×10−19 C) and Vis the potential difference (100 V). The

kinetic energy of the electron is equal to the potential energy gained, so:

K.E. =qV = (1.6×10−19 C)(100 V)

Step 2: Calculate the velocity of the electron. The kinetic energy is also

equal to 1

2mv2, where mis the mass of the electron (9.11 ×10−31 kg) and vis

the velocity. Equating the two expressions for kinetic energy:

1

2mv2=qV

v=r2qV

m

Step 3: Calculate the de Broglie wavelength using the velocity. The de

Broglie wavelength (λ) is given by λ=h

mv , where his the Planck constant

(6.63 ×10−34 J s). Substitute the calculated velocity into the formula for de

Broglie wavelength:

λ=h

mv =6.63 ×10−34 J s

(9.11 ×10−31 kg)(q2(1.6×10−19 C)(100 V)

9.11×10−31 kg )

6

Question 7

Question

A particle of mass mmoves with a velocity 4 times the velocity of another

particle of mass 3m. If the de Broglie wavelength of the first particle is λ, what

is the de Broglie wavelength of the second particle?

Solution

Given: Velocity of first particle = 4v

Velocity of second particle = v

Mass of first particle = m

Mass of second particle = 3m

De Broglie wavelength of first particle = λ

Let the de Broglie wavelength of the second particle be λ′.

According to de Broglie relation, the de Broglie wavelength is given by:

λ=h

pand λ′=h

p′

where pand p′are the momenta of the first and second particles respectively.

Step 1: Find the momentum of the first particle:

The momentum pof a particle is given by p=mv. Therefore, the momentum

of the first particle is:

p=m·4v= 4mv

Step 2: Find the momentum of the second particle:

The momentum p′of the second particle is given by p′= 3mv.

Step 3: Find the de Broglie wavelength of the second particle:

Substitute the momentum expressions into the de Broglie wavelength for-

mula to find the de Broglie wavelength of the second particle:

λ′=h

p′=h

3mv

Therefore, the de Broglie wavelength of the second particle is h

3mv .

Question 8

Question

An electron is accelerated from rest through a potential difference of 100 V.

Determine the de Broglie wavelength associated with the electron after this

acceleration.

7

Solution

Step 1: Find the kinetic energy of the electron after acceleration. Given that

the potential difference is 100 V and the charge of an electron is −1.6×10−19

C, the kinetic energy can be calculated using the equation:

K.E. =q·∆V= (1.6×10−19 C) ×(100 V)

Step 2: Calculate the kinetic energy.

K.E. = 1.6×10−17 J

Step 3: Use the kinetic energy to find the de Broglie wavelength. The de

Broglie wavelength is given by:

λ=h

p=h

√2mK.E.

where his the Planck constant (6.626 ×10−34 Js), mis the mass of the electron

(9.11 ×10−31 kg), and K.E. is the kinetic energy.

Step 4: Substitute the values to find the de Broglie wavelength.

λ=6.626 ×10−34 Js

p2×9.11 ×10−31 kg ×1.6×10−17 J

Step 5: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 Js

p2×9.11 ×10−31 kg ×1.6×10−17 J

λ≈1.21 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron after ac-

celeration through a potential difference of 100 V is approximately 1.21 ×10−10

m.

Question 9

Question

An electron is accelerated through a potential difference of 1000 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the formula K=eV ,

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 2: The kinetic energy of the electron can be calculated as:

K= (1.6×10−19 C)(1000 V)

8

Step 3: Simply the expression to find the kinetic energy:

K= 1.6×10−16 J

Step 4: Using the formula for the de Broglie wavelength λ=h

p, where his

Planck’s constant (6.63 ×10−34 J s) and pis the momentum of the electron.

Step 5: The momentum of the electron is given by p=√2mK, where mis

the mass of the electron (9.11 ×10−31 kg).

Step 6: Calculate the momentum of the electron:

p=p2(9.11 ×10−31 kg)(1.6×10−16 J)

Step 7: Simplify to find the momentum:

p≈3.02 ×10−24 kg m/s

Step 8: Finally, determine the de Broglie wavelength:

λ=6.63 ×10−34 J s

3.02 ×10−24 kg m/s

Step 9: Simplify to find the de Broglie wavelength:

λ≈2.20 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 1000 V is approximately 2.20 ×10−10 m.

Question 10

Question

A beam of electrons with kinetic energy Eis directed toward a single slit of

width a. Calculate the de Broglie wavelength of the electrons after passing

through the slit.

Solution

Step 1: Calculate the momentum of the electrons using their kinetic energy.

The momentum of an electron is given by the formula p=√2mE, where mis

the mass of the electron and Eis its kinetic energy.

Step 2: Calculate the de Broglie wavelength of the electrons. The de Broglie

wavelength is given by the formula λ=h

p, where his the Planck constant.

Therefore, the de Broglie wavelength of the electrons after passing through

the slit is λ=h

√2mE .

9

Question 11

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron after acceleration.

Solution

Step 1: First, we need to calculate the kinetic energy gained by the electron

when accelerated through a potential difference of 100 V. The kinetic energy of

the electron can be calculated using the equation:

KE =qV

where KE is the kinetic energy, qis the charge of the electron, and Vis the

potential difference. Given that the charge of an electron q=−1.6×10−19 C,

and the potential difference V= 100 V, we can substitute these values into the

formula:

KE = (−1.6×10−19 C)(100 V)

KE =−1.6×10−17 J

Step 2: Next, we convert the kinetic energy into the momentum of the

electron using the equation:

KE =1

2mv2=p2

2m

where mis the mass of the electron and pis the momentum. Rearranging the

formula to solve for momentum, we get:

p=√2mKE

Given that the mass of an electron m= 9.11 ×10−31 kg, we can substitute the

values of mass and kinetic energy into the formula:

p=p2(9.11 ×10−31 kg)(1.6×10−17 J)

p≈4.88 ×10−24 kg m/s

Step 3: Finally, we can calculate the de Broglie wavelength of the electron

using the formula:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),

and pis the momentum of the electron. Substitute the values of Planck constant

and momentum into the formula to find the de Broglie wavelength:

λ=6.626 ×10−34 m2kg/s

4.88 ×10−24 kg m/s

10

λ≈1.36 ×10−10 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 100 V is approximately 1.36 ×10−10 m.

Question 12

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula KE =eV ,

where eis the elementary charge and Vis the potential difference. Step 2:

Determine the momentum of the electron using the formula p=√2mKE, where

mis the mass of the electron. Step 3: Calculate the de Broglie wavelength using

the formula λ=h

p, where his the Planck constant. Given data: V= 200 V,

e= 1.6×10−19 C, m= 9.11 ×10−31 kg, and h= 6.626 ×10−34 J·s.

Step 1: Find the kinetic energy of the electron.

KE =eV = (1.6×10−19C)(200V)=3.2×10−17J

Step 2: Determine the momentum of the electron.

p=√2mKE =p2(9.11 ×10−31kg)(3.2×10−17J)≈1.77 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength.

λ=h

p=6.626 ×10−34J·s

1.77 ×10−24 kg m/s ≈3.74 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 3.74 ×10−10 m.

Question 13

Question

An electron is accelerated through a potential difference of 150 V. Determine:

(a) The de Broglie wavelength associated with the electron. (b) The velocity of

the electron.

11

Solution

(a) Let’s first determine the de Broglie wavelength associated with the electron

using the formula:

λ=h

p

where λ= de Broglie wavelength (m), h= Planck’s constant = 6.626×10−34 Js,

p= momentum of the particle.

We know that momentum (p) can be calculated using the formula:

p=√2mE

where m= mass of the electron = 9.11 ×10−31 kg, E= kinetic energy of the

electron.

Given that the potential difference is 150 V, we can calculate the kinetic

energy using the formula:

E=qV

where q= charge of an electron = 1.6×10−19 C, V= potential difference =

150 V.

Step 1: Calculate the kinetic energy of the electron.

E=qV = (1.6×10−19 C)(150 V)

E= 2.4×10−17 J

Step 2: Calculate the momentum of the electron.

p=√2mE =p2(9.11 ×10−31 kg)(2.4×10−17 J)

p≈5.44 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength.

λ=h

p=6.626 ×10−34 Js

5.44 ×10−24 kg m/s

λ≈1.22 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron is approx-

imately 1.22 ×10−10 m.

(b) We can now determine the velocity of the electron using the de Broglie

wavelength:

v=h

mλ

where v= velocity of the electron.

Step 1: Calculate the velocity of the electron.

v=h

mλ =6.626 ×10−34 Js

(9.11 ×10−31 kg)(1.22 ×10−10 m)

v≈5.39 ×106m/s

Therefore, the velocity of the electron is approximately 5.39 ×106m/s.

12

Question 14

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula, which relates the wave-

length of a particle to its momentum. The formula is given by:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J·s),

and pis the momentum of the particle.

Step 2: First, we need to find the momentum of the electron. The kinetic

energy of the electron can be calculated using the formula:

K.E. =qV

where qis the charge of the electron (−1.6×10−19 C) and Vis the potential

difference (200 V in this case).

Step 3: Substituting the values into the formula for kinetic energy:

K.E. = (−1.6×10−19 C)(200 V) = −3.2×10−17 J

Step 4: The kinetic energy can also be expressed in terms of momentum as:

K.E. =p2

2m

where mis the mass of the electron.

Step 5: Rearranging the formula to solve for momentum, we get:

p=√2mK.E.

Step 6: Substituting the values into the formula for momentum:

p=p2×9.11 ×10−31 kg ×3.2×10−17 J≈4.44 ×10−24 kg ·m/s

Step 7: Finally, substituting the momentum into the de Broglie wavelength

formula:

λ=6.626 ×10−34 J·s

4.44 ×10−24 kg ·m/s ≈1.49 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 1.49 ×10−10 m.

13

Question 15

Question

A beam of electrons with kinetic energy 500 eV is incident on a crystal. If

the crystal diffracts the electrons at an angle of 30◦, what is the de Broglie

wavelength of the electrons?

Solution

Step 1: We first find the velocity of the electrons using the kinetic energy formula

KE =1

2mev2, where meis the mass of an electron (9.11 ×10−31 kg).

Step 2: Rearranging the kinetic energy formula gives us v=q2×500 eV×1.6×10−19 C/eV

9.11×10−31 kg .

Converting the kinetic energy to joules and solving for velocity gives v≈

2.19 ×106m/s.

Step 3: Using the definition of de Broglie wavelength λ=h

p, where his the

Planck constant (6.626 ×10−34 J s) and pis the momentum of the electron.

Step 4: The momentum of an electron can be calculated as p=mev, and

substituting the values gives p= (9.11 ×10−31 kg) ×(2.19 ×106m/s). Solving

for momentum gives p≈2.00 ×10−24 kg m/s.

Step 5: Substituting the momentum into the de Broglie wavelength formula

gives λ=6.626×10−34 J s

2.00×10−24 kg m/s . Calculating the de Broglie wavelength gives λ≈

3.31 ×10−12 m.

Therefore, the de Broglie wavelength of the electrons is approximately 3.31×

10−12 m.

Question 16

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron in meters.

Solution

Step 1: Determine the kinetic energy of the electron using the formula K=eV ,

where eis the elementary charge and Vis the potential difference. Step 2:

Use the kinetic energy of the electron to find its momentum using the formula

p=√2mK, where mis the mass of the electron and Kis the kinetic energy.

Step 3: Calculate the de Broglie wavelength using the formula λ=h

p, where h

is the Planck constant and pis the momentum of the electron. Step 4: Convert

the de Broglie wavelength to meters.

Now let’s proceed with the calculations.

Step 1: Determine the kinetic energy of the electron. Given: Potential

difference, V= 100 V Elementary charge, e= 1.6×10−19 C

14

Using K=eV :K= (1.6×10−19 C)(100 V) K= 1.6×10−17 J

Step 2: Calculate the momentum of the electron. Given: Mass of electron,

m= 9.109 ×10−31 kg

Using p=√2mK:p=p2(9.109 ×10−31 kg)(1.6×10−17 J) p≈4.740 ×

10−24 kg m/s

Step 3: Calculate the de Broglie wavelength. Given: Planck constant,

h= 6.626 ×10−34 m2kg/s

Using λ=h

p:λ=6.626×10−34 m2kg/s

4.740×10−24 kg m/s λ≈1.399 ×10−10 m

Step 4: Convert the de Broglie wavelength to meters. The de Broglie

wavelength of the electron is approximately 1.399 ×10−10 meters.

Question 17

Question

An electron is accelerated through a potential difference of 100 V. Calculate

the de Broglie wavelength of the electron after being accelerated. (Mass of an

electron = 9.11 ×10−31 kg, charge of an electron = 1.60 ×10−19 C, Planck’s

constant = 6.63 ×10−34 J·s)

Solution

Step 1: Calculate the kinetic energy of the electron. The kinetic energy of the

electron can be calculated using the formula:

KE =qV

where: KE = kinetic energy (J), q= charge of the electron (C), V= potential

difference (V).

Substitute the given values:

KE = (1.60 ×10−19 C)(100 V) = 1.60 ×10−17 J

Step 2: Calculate the velocity of the electron. The kinetic energy of the

electron can also be expressed in terms of its velocity vusing the formula:

KE =1

2mv2

where: m= mass of the electron (kg), v= velocity of the electron (m/s).

Equating the two expressions for kinetic energy:

1

2mv2= 1.60 ×10−17

Solving for v:

v=r2×1.60 ×10−17

m

15

v=r2×1.60 ×10−17

9.11 ×10−31

v=r3.20 ×10−17

9.11 ×10−31

v≈√3.514 ×107

v≈5.92 ×103m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength λof an electron is given by the formula:

λ=h

p

where: h= Planck’s constant (J·s), p= momentum of the electron (kg·m/s).

The momentum pof the electron is given by:

p=mv

p= (9.11 ×10−31 kg)(5.92 ×103m/s)

p≈5.40 ×10−27 kg ·m/s

Substitute the values into the de Broglie wavelength formula:

λ=6.63 ×10−34 J·s

5.40 ×10−27 kg·m/s

λ≈6.63 ×10−34

5.40 ×107

λ≈1.23 ×10−10 m

Therefore, the de Broglie wavelength of the electron after being accelerated

through a potential difference of 100 V is approximately 1.23 ×10−10 m.

Question 18

Question

An electron is accelerated through a potential difference of 500 V. What is the

de Broglie wavelength associated with this electron?

16

Solution

Step 1: Use the formula for de Broglie wavelength:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),

and pis the momentum of the particle.

Step 2: First, find the kinetic energy of the electron using the formula:

kinetic energy (KE) = charge ×potential difference

Given that the charge of an electron e= 1.6×10−19 C:

KE = e×potential difference

Step 3: Calculate the velocity of the electron using the formula for kinetic

energy:

KE = 1

2mv2

where mis the mass of the electron (9.11 ×10−31 kg) and vis the velocity of

the electron.

Step 4: Calculate the momentum pof the electron using its mass and veloc-

ity:

p=m×v

Step 5: Substitute the values of Planck’s constant hand momentum pinto

the de Broglie wavelength equation to find the de Broglie wavelength λ.

Question 19

Question

An electron is accelerated through a potential difference of 150 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula for particles:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J·s),

and pis the momentum of the particle.

Step 2: To find the momentum, we can use the formula for the momentum

of a particle:

p=√2mE

17

where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy

of the electron.

Step 3: The kinetic energy of the electron can be found using the formula:

E=qV

where qis the charge of the electron (−1.6×10−19 C) and Vis the potential

difference it is accelerated through (150 V).

Step 4: Substituting the kinetic energy into the momentum formula gives:

p=p2m(qV )

Step 5: Now, substituting the momentum into the de Broglie wavelength

formula gives:

λ=h

p2m(qV )

Step 6: Plugging in the values (h= 6.626 ×10−34 J·s, m= 9.11 ×10−31 kg,

q=−1.6×10−19 C, and V= 150 V) and calculating gives:

λ=6.626 ×10−34 J·s

p2(9.11 ×10−31 kg)(−1.6×10−19 C)(150 V)

Step 7: Solving for λgives the de Broglie wavelength associated with the

electron accelerated through the potential difference of 150 V.

Question 20

Question

An electron is accelerated through a potential difference of 120 V. Determine

the de Broglie wavelength of the electron after passing through the potential

difference.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

Given that the potential difference is 120 V, we can calculate the kinetic energy

using the formula K.E. =qV , where qis the charge of the electron (1.6 ×10−19

C) and Vis the potential difference.

Plugging in the values, we get:

K.E. = (1.6×10−19 C) ×(120 V) = 1.92 ×10−17 J

Step 2: Use the kinetic energy to find the momentum of the electron. The

kinetic energy of the electron can be related to its momentum using the formula

K.E. =1

2mv2=p2

2m, where mis the mass of the electron and vis its velocity.

18

Therefore, we get:

p2= 2m×K.E.

p=√2m×K.E.

Substitute the values of m(mass of electron = 9.11 ×10−31 kg) and K.E.:

p=p2×9.11 ×10−31 kg ×1.92 ×10−17 J=3.12 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength (λ) is related to the momentum of the electron by the formula λ=h

p,

where his the Planck constant (6.626 ×10−34 J s).

Substitute the values to find the de Broglie wavelength:

λ=6.626 ×10−34 J s

3.12 ×10−24 kg m/s = 2.12 ×10−10 m

Therefore, the de Broglie wavelength of the electron after passing through

the potential difference is 2.12 ×10−10 meters.

Question 21

Question

An electron is accelerated through a potential difference of 250 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula Ek=qV ,

where qis the electron charge and Vis the potential difference.

Given: V= 250 V (potential difference)

Elementary charge, e= 1.6×10−19 C

Ek=eV = (1.6×10−19 C)(250 V)

Ek= 4 ×10−17 J

Step 2: Use the formula for kinetic energy to find the momentum of the

electron.

Ek=p2

2m(where pis momentum and mis mass)

p=p2mEk

Mass of electron, m= 9.11 ×10−31 kg

p=p2(9.11 ×10−31 kg)(4 ×10−17 J)

p≈6.09 ×10−24 kg m/s

19

Step 3: Calculate the de Broglie wavelength using the formula λ=h

p,

where his the Planck constant.

h= 6.63 ×10−34 J s

λ=6.63 ×10−34 J s

6.09 ×10−24 kg m/s

λ≈1.09 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 250 V is approximately 1.09 ×10−10 m.

Question 22

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula to find the wavelength

associated with the electron:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J

·s), and pis the momentum of the particle.

Step 2: First, we need to find the momentum of the electron. The kinetic

energy of the electron can be calculated using the potential difference it was

accelerated through:

K.E. =e·V

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 3: Substitute the given values into the equation to find the kinetic

energy of the electron:

K.E. = (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 4: Next, we use the relationship between kinetic energy and momentum

to find the momentum of the electron:

K.E. =p2

2m

where mis the mass of the electron (9.11 ×10−31 kg).

Step 5: Solve for the momentum p:

p=√2m·K.E. =p2×9.11 ×10−31 kg ×1.6×10−17 J≈2.20 ×10−24 kg m/s

20

Step 6: Finally, substitute the momentum into the de Broglie wavelength

formula to find the wavelength:

λ=6.626 ×10−34 J s

2.20 ×10−24 kg m/s ≈3.01 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 100 V is approximately 3.01 ×10−10 m.

Question 23

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference

given. Step 2: Use the kinetic energy to find the velocity of the electron. Step

3: Use the velocity to calculate the de Broglie wavelength of the electron.

Step 1: Calculate the kinetic energy of the electron. Using the formula for

kinetic energy, K.E. =qV , where qis the charge of an electron (1.6×10−19 C)

and Vis the potential difference (200 V):

K.E. = (1.6×10−19 C)(200 V) = 3.2×10−17 J

Step 2: Find the velocity of the electron. Since the kinetic energy is equal

to 1

2mv2, where mis the mass of the electron (9.11 ×10−31 kg), we can solve

for the velocity v:

3.2×10−17 J = 1

2(9.11 ×10−31 kg)v2

v=s2×3.2×10−17 J

9.11 ×10−31 kg ≈6.09 ×106m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength is given by λ=h

p, where his the Planck constant (6.626 ×10−34

J·s) and pis the momentum of the electron (m·v):

λ=6.626 ×10−34 J·s

9.11 ×10−31 kg ·6.09 ×106m/s ≈1.36 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 1.36 ×10−10 m.

21

Question 24

Question

A proton is moving with a velocity of 2.5×106m/s. Calculate the de Broglie

wavelength associated with the proton’s motion.

Solution

Step 1: Recall the de Broglie wavelength formula:

λ=h

p

where: - λis the de Broglie wavelength, - his the Planck constant (6.626×10−34

J·s), and - pis the momentum of the particle.

Step 2: Calculate the momentum of the proton using the equation p=mv,

where mis the mass of the proton and vis its velocity.

p= (1.67 ×10−27 kg) ×(2.5×106m/s)

Step 3: Calculate the momentum of the proton.

p= 4.175 ×10−21 kg m/s

Step 4: Substitute the momentum into the de Broglie wavelength formula

to get the de Broglie wavelength for the proton.

λ=6.626 ×10−34 J·s

4.175 ×10−21 kg m/s

Step 5: Calculate the de Broglie wavelength.

λ≈1.587 ×10−13 m

Therefore, the de Broglie wavelength associated with the proton’s motion is

approximately 1.587 ×10−13 m.

Question 25

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength associated with this electron.

22

Solution

Step 1: Determine the kinetic energy of the electron using the potential differ-

ence. Step 2: Apply the de Broglie wavelength formula to find the wavelength

associated with the electron.

Step 1: The kinetic energy of the electron can be calculated as

K.E. =qV

where qis the charge of the electron and Vis the potential difference. Given that

the potential difference is 200 V and the charge of an electron is −1.6×10−19

C, we have

K.E. = (−1.6×10−19 C) ×(200 V) = −3.2×10−17 J

Step 2: The de Broglie wavelength of the electron is given by

λ=h

p

where his the Planck’s constant (6.626×10−34 m2kg/s) and pis the momentum

of the electron. The momentum of the electron can be calculated using the

kinetic energy:

p=√2mK.E.

where mis the mass of the electron. Substitute the values in to find the de

Broglie wavelength associated with the electron.

Question 26

Question

Two particles, one with mass m1and another with mass m2, are moving with

the same velocity. Calculate the ratio of their de Broglie wavelengths, λ1

λ2.

Solution

1. The de Broglie wavelength of a particle is given by:

λ=h

p

where his the Planck constant and pis the momentum of the particle.

2. The momentum pof a particle is given by:

p=mv

where mis the mass of the particle and vis its velocity.

23

3. Therefore, the de Broglie wavelength λof a particle with mass mand

velocity vcan be written as:

λ=h

mv

4. Let’s calculate the de Broglie wavelengths λ1and λ2for particles with

masses m1and m2moving at the same velocity.

For particle 1 with mass m1:

λ1=h

m1v

For particle 2 with mass m2:

λ2=h

m2v

5. Now, we’ll find the ratio λ1

λ2:

λ1

λ2

=

h

m1v

h

m2v

=m2

m1

6. Therefore, the ratio of the de Broglie wavelengths of the two particles is

m2

m1

.

Question 27

Question

An electron is accelerated through a potential difference of 100 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Use the formula relating the de Broglie wavelength (λ) with the mo-

mentum (p) of a particle:

λ=h

p

where his the Planck constant (6.63 ×10−34 m2kg/s).

Step 2: Find the momentum of the electron using its kinetic energy: The

kinetic energy of an electron accelerated through a potential difference is given

by:

K.E. =eV

24

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Given that the potential difference (V) is 100 V, the kinetic energy of the

electron is:

K.E. =e×100 = 1.6×10−17 J

Step 3: Use the relationship between kinetic energy and momentum to find

the momentum p:

K.E. =p2

2m

where mis the mass of the electron (9.11 ×10−31 kg).

Substitute the values to find the momentum:

1.6×10−17 =p2

2×9.11 ×10−31

p=p2×9.11 ×10−31 ×1.6×10−17

Step 4: Calculate the de Broglie wavelength using the momentum obtained:

Now substitute the momentum into the formula for de Broglie wavelength:

λ=6.63 ×10−34

√2×9.11 ×10−31 ×1.6×10−17

λ≈6.63 ×10−34

2.14 ×10−24 ≈3.1×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 100 V is approximately 3.1×10−10 m.

Question 28

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength of the electron after acceleration. (Take the charge of an

electron, e, to be 1.6×10−19 C.)

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

The kinetic energy gained by the electron is equal to the work done on it by the

potential difference:

Kinetic energy (KE) = Charge ×Potential difference

KE = e×V

KE = 1.6×10−19 C×200 V

25

KE = 3.2×10−17 J

Step 2: Use the kinetic energy of the electron to find its momentum. The

momentum of the electron is given by:

Momentum = p2m×Kinetic energy

where mis the mass of the electron. The mass of an electron is 9.11 ×10−31 kg.

Momentum = p2×9.11 ×10−31 kg ×3.2×10−17 J

Momentum ≈2.19 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength using the momentum of the

electron. The de Broglie wavelength is given by:

λ=h

p

where his the Planck constant, 6.63 ×10−34 J s.

λ=6.63 ×10−34 J s

2.19 ×10−24 kg m/s

λ≈3.03 ×10−11 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 200 V is approximately 3.03 ×10−11 m.

Question 29

Question

A particle of mass m= 5.0×10−27 kg moves with a speed of 4.0×105m/s.

Calculate the de Broglie wavelength associated with this particle.

Solution

Step 1: Calculate the momentum of the particle using the formula p=mv,

where mis the mass and vis the speed.

p= (5.0×10−27 kg)(4.0×105m/s)

p= 2.0×10−21 kg m/s

Step 2: Calculate the de Broglie wavelength using the formula λ=h

p, where

his the Planck constant.

λ=6.63 ×10−34 J s

2.0×10−21 kg m/s

λ= 3.32 ×10−13 m

Therefore, the de Broglie wavelength associated with the particle is 3.32 ×

10−13 meters.

26

Question 30

Question

An electron is accelerated from rest through a potential difference of 120 V.

Determine the de Broglie wavelength associated with the electron after acceler-

ation.

Solution

Step 1: We know that the kinetic energy of the electron after acceleration

through a potential difference of Vis given by the equation:

KE =eV

where eis the elementary charge and Vis the potential difference.

Step 2: Substituting the given values, we find the kinetic energy of the

electron:

KE = (1.6×10−19 C)(120 V)

KE = 1.92 ×10−17 J

Step 3: The de Broglie wavelength of a particle is given by the equation:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J s),

and pis the momentum of the particle.

Step 4: The momentum of the electron can be calculated using its kinetic

energy:

p=p2meKE

where meis the mass of the electron (9.11 ×10−31 kg).

Step 5: Substituting the values and calculating the momentum:

p=p2(9.11 ×10−31 kg)(1.92 ×10−17 J)

p≈4.38 ×10−24 kg m/s

Step 6: Finally, we can determine the de Broglie wavelength of the electron:

λ=6.626 ×10−34 J s

4.38 ×10−24 kg m/s

λ≈1.51 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron after ac-

celeration is approximately 1.51 ×10−10 m.

27

Question 31

Question

A particle of mass mmoves with a velocity v. By applying the de Broglie

hypothesis, calculate the de Broglie wavelength associated with this particle.

Solution

Step 1: According to the de Broglie hypothesis, the wavelength (λ) associated

with a particle of momentum pis given by:

λ=h

p

where his the Planck constant.

Step 2: The momentum of a particle of mass mmoving with velocity vis

given by:

p=mv

Step 3: Substitute the expression for momentum into the de Broglie wave-

length formula:

λ=h

mv

Step 4: Therefore, the de Broglie wavelength associated with a particle of

mass mmoving with velocity vis h/(mv).

Question 32

Question

An electron is accelerated from rest through a potential difference of 500 V.

Calculate the de Broglie wavelength of the electron after acceleration.

Solution

Step 1: First, we need to calculate the kinetic energy (K.E.) of the electron after

acceleration using the formula K.E. =eV , where eis the charge of an electron

and Vis the potential difference. Given e= 1.6×10−19 C and V= 500 V, we

substitute these values into the formula:

K.E. = (1.6×10−19 C) ×(500 V)

K.E. = 8 ×10−17 J

Step 2: Next, we use the kinetic energy of the electron to find its velocity

vusing the formula K.E. =1

2mv2, where mis the mass of the electron. The

28

mass of an electron is 9.11 ×10−31 kg. Substituting the values into the formula,

we get:

8×10−17 J = 1

2×(9.11 ×10−31 kg) ×v2

v=s2×8×10−17 J

9.11 ×10−31 kg

v≈2.75 ×106m/s

Step 3: Finally, we can calculate the de Broglie wavelength λof the electron

using the formula λ=h

p, where his the Planck constant and pis the momentum

of the electron. The momentum pcan be calculated as p=mv. Substituting

the values into the formula, we get:

λ=h

mv =6.626 ×10−34 J s

(9.11 ×10−31 kg) ×(2.75 ×106m/s)

λ≈2.44 ×10−12 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 500 V is approximately 2.44 ×10−12 m.

Question 33

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Recall the de Broglie wavelength formula: λ=h

p, where his the Planck

constant (6.63 ×10−34 J·s) and pis the momentum of the particle.

Step 2: To find the momentum p, we can use the relation between kinetic

energy, charge, and potential difference: K.E. =qV , where qis the charge of

the electron and Vis the potential difference.

Step 3: The kinetic energy of the electron can be calculated as K.E. =eV ,

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 4: Substituting K.E. =eV into the expression for momentum p=

√2mK.E., where mis the mass of the electron, we have p=√2meV .

Step 5: Now, we have the momentum pin terms of the potential difference

V. Substituting this into the de Broglie wavelength formula λ=h

p, we can find

the de Broglie wavelength associated with the electron.

29

Step 6: Substituting p=√2meV into λ=h

pgives:

λ=h

√2meV

Step 7: Plugging in the values for h,m,e, and V:

λ=6.63 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−19 C×200 V

Step 8: Now, calculate the de Broglie wavelength using the above formula.

Step 9: The de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately λ≈1.23 ×10−10 m.

Question 34

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength associated with the electron.

Solution

Step 1: Determine the kinetic energy of the electron using the potential differ-

ence. Given that the potential difference is 100 V and the charge of an electron

is −1.6×10−19 C, we can calculate the kinetic energy as:

K=qV =−1.6×10−19 C×100 V

Step 2: Calculate the de Broglie wavelength using the formula λ=h

p. Since

the kinetic energy is related to the momentum pby K=p2

2m, we can express p

in terms of Kand m:

p=√2mK

where mis the mass of an electron. We can then substitute this expression for

pinto the formula for the de Broglie wavelength:

λ=h

√2mK

Step 3: Substitute the known values and calculate the de Broglie wavelength.

Given that Planck’s constant h= 6.626 ×10−34 J s and the mass of an electron

m= 9.11 ×10−31 kg, we can substitute these values along with the calculated

kinetic energy into the formula:

λ=6.626 ×10−34 J s

p2×9.11 ×10−31 kg ×(−1.6×10−19 C×100 V)

Step 4: Simplify and calculate the de Broglie wavelength. After simplify-

ing the expression and performing the calculations, we can find the de Broglie

wavelength associated with the electron.

30

Question 35

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Recall the de Broglie wavelength formula: The de Broglie wavelength

(λ) of a particle is given by:

λ=h

p

where his the Planck constant (6.626 ×10−34 J·s) and pis the momentum of

the particle.

Step 2: Find the momentum of the electron: The momentum of an electron

can be calculated using the equation p=√2meE, where meis the mass of an

electron (9.11 ×10−31 kg) and Eis the kinetic energy of the electron.

Step 3: Determine the kinetic energy of the electron: The kinetic energy of

the electron can be calculated using the equation E=qV , where qis the charge

of an electron (−1.6×10−19 C) and Vis the potential difference.

Step 4: Calculate the kinetic energy: Substitute the values q=−1.6×10−19

C and V= 200 V into the equation E=qV to find the kinetic energy of the

electron.

Step 5: Calculate the momentum: Substitute the calculated kinetic energy

into the equation p=√2meEto find the momentum of the electron.

Step 6: Find the de Broglie wavelength: Finally, substitute the calculated

momentum into the equation λ=h

pto determine the de Broglie wavelength

associated with the electron.

31

5. Rearranging the equation gives:

v=r2E

m=s2× −1.6×10−17 J

9.11 ×10−31 kg

6. Calculating the velocity gives:

v≈5.92 ×106m/s

7. The de Broglie wavelength (λ) can be calculated using the equation:

λ=h

mv

where his the Planck constant.

8. Substituting the values of h,m, and v, we get:

λ=6.63 ×10−34 J s

(9.11 ×10−31 kg) ×(5.92 ×106m/s)

9. Calculating the de Broglie wavelength gives:

λ≈1.22 ×10−10 m

Question 2

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron. (Hint: The energy of the electron can be

related to its de Broglie wavelength through the formula E=p2

2m, where Eis

the kinetic energy, pis the momentum, and mis the mass of the electron.)

Solution

Step 1: We know that the kinetic energy of the electron is given by E=qV ,

where qis the charge of the electron and Vis the potential difference. Since the

charge of an electron is 1.6×10−19 C and the potential difference is 100 V, the

kinetic energy of the electron is

E= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 2: The kinetic energy of the electron can also be expressed in terms

of momentum as E=p2

2m, where pis the momentum and mis the mass of the

electron. Therefore, we have

p2

2m= 1.6×10−17 J

2

p2= 3.2×10−17 ×m

Step 3: The de Broglie wavelength of the electron is given by λ=h

p, where

his Planck’s constant. Substituting p=√3.2×10−17 ×minto the de Broglie

wavelength formula, we get

λ=h

√3.2×10−17 ×m

Step 4: The mass of an electron is 9.11 ×10−31 kg. Substituting this value

and h= 6.63 ×10−34 J s into the de Broglie wavelength formula, we find

λ=6.63 ×10−34

√3.2×10−17 ×9.11 ×10−31

λ≈7.27 ×10−11 m

Therefore, the de Broglie wavelength of the electron accelerated through a

potential difference of 100 V is approximately 7.27 ×10−11 m.

Question 3

Question

An electron with a kinetic energy of 200 eV is moving in vacuum. Find the de

Broglie wavelength associated with this electron.

Solution

Step 1: Convert the kinetic energy from electron-volts to joules.

1 eV = 1.6×10−19 J

So, the kinetic energy of the electron in joules is:

E= 200 eV ×1.6×10−19 J/eV = 3.2×10−17 J

Step 2: Use the de Broglie wavelength formula to find the wavelength asso-

ciated with the electron. The de Broglie wavelength is given by:

λ=h

p

where λ= wavelength, h= Planck’s constant (6.626 ×10−34 J s), p= momen-

tum = √2mE,m= mass of the electron (9.11 ×10−31 kg), E= kinetic energy

of the electron.

Step 3: Calculate the momentum of the electron.

p=p2×9.11 ×10−31 kg ×3.2×10−17 J

3

p=q5.82 ×10−16 kg m2s−2

p= 7.63 ×10−8kg m/s

Step 4: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 J s

7.63 ×10−8kg m/s

λ≈8.68 ×10−11 m

Therefore, the de Broglie wavelength associated with the electron with a

kinetic energy of 200 eV is approximately 8.68 ×10−11 meters.

Question 4

Question

A beam of electrons with kinetic energy 1 keV is incident on a crystal lattice.

Assuming that the electrons behave as de Broglie waves, calculate the de Broglie

wavelength of the electrons.

Solution

Step 1: Calculate the momentum of the electrons. The momentum of an

electron can be calculated using the formula:

p=√2mE

where mis the mass of the electron and Eis the kinetic energy.

Given:

E= 1 keV = 1.6×10−16 J

m= 9.11 ×10−31 kg

Substitute the given values into the formula:

p=p2×9.11 ×10−31 ×1.6×10−16

p=p2×9.11 ×1.6×10−47

p=p2×14.576 ×10−47

p=p29.152 ×10−47

p= 5.401 ×10−24 kg m/s

Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength of

an electron is given by:

λ=h

p

4

where his the Planck constant.

The Planck constant is h= 6.626 ×10−34 m2kg/s.

Substitute the values of hand pinto the formula:

λ=6.626 ×10−34

5.401 ×10−24

λ=6.626

5.401 ×10−10

λ= 1.227 ×10−10 m

Therefore, the de Broglie wavelength of the electrons is 1.227 ×10−10 m.

Question 5

Question

An electron is accelerated through a potential difference of 120 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference

provided. Step 2: Use the kinetic energy to find the velocity of the electron. Step

3: Apply the de Broglie wavelength formula to find the wavelength associated

with the electron.

Step 1: Calculate the kinetic energy of the electron. The kinetic energy

(KE) of an electron accelerated through a potential difference ∆Vis given by:

KE =e·∆V

where eis the elementary charge (1.6×10−19 C) and ∆Vis the potential

difference (120 V).

KE = (1.6×10−19 C) ×(120 V) = 1.92 ×10−17 J

Step 2: Calculate the velocity of the electron. The kinetic energy of the

electron can be expressed in terms of its velocity vas:

KE =1

2mv2

where mis the mass of the electron (9.11 ×10−31 kg). Solving for v:

v=r2KE

m=r2×1.92 ×10−17

9.11 ×10−31 ≈4.08 ×106m/s

5

Step 3: Calculate the de Broglie wavelength. The de Broglie wavelength

(λ) of a particle is given by:

λ=h

p

where his the Planck constant (6.63 ×10−34 J·s) and pis the momentum of

the electron. The momentum of the electron is given by p=mv, where mis

the mass of the electron and vis the velocity.

p= (9.11 ×10−31 kg) ×(4.08 ×106m/s) ≈3.72 ×10−24 kg ·m/s

λ=6.63 ×10−34

3.72 ×10−24 ≈1.78 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 120 V is approximately 1.78 ×10−10 m.

Question 6

Question

An electron is accelerated through a potential difference of 100 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

The potential energy gained by the electron is given by qV , where qis the charge

of the electron (1.6 ×10−19 C) and Vis the potential difference (100 V). The

kinetic energy of the electron is equal to the potential energy gained, so:

K.E. =qV = (1.6×10−19 C)(100 V)

Step 2: Calculate the velocity of the electron. The kinetic energy is also

equal to 1

2mv2, where mis the mass of the electron (9.11 ×10−31 kg) and vis

the velocity. Equating the two expressions for kinetic energy:

1

2mv2=qV

v=r2qV

m

Step 3: Calculate the de Broglie wavelength using the velocity. The de

Broglie wavelength (λ) is given by λ=h

mv , where his the Planck constant

(6.63 ×10−34 J s). Substitute the calculated velocity into the formula for de

Broglie wavelength:

λ=h

mv =6.63 ×10−34 J s

(9.11 ×10−31 kg)(q2(1.6×10−19 C)(100 V)

9.11×10−31 kg )

6

Question 7

Question

A particle of mass mmoves with a velocity 4 times the velocity of another

particle of mass 3m. If the de Broglie wavelength of the first particle is λ, what

is the de Broglie wavelength of the second particle?

Solution

Given: Velocity of first particle = 4v

Velocity of second particle = v

Mass of first particle = m

Mass of second particle = 3m

De Broglie wavelength of first particle = λ

Let the de Broglie wavelength of the second particle be λ′.

According to de Broglie relation, the de Broglie wavelength is given by:

λ=h

pand λ′=h

p′

where pand p′are the momenta of the first and second particles respectively.

Step 1: Find the momentum of the first particle:

The momentum pof a particle is given by p=mv. Therefore, the momentum

of the first particle is:

p=m·4v= 4mv

Step 2: Find the momentum of the second particle:

The momentum p′of the second particle is given by p′= 3mv.

Step 3: Find the de Broglie wavelength of the second particle:

Substitute the momentum expressions into the de Broglie wavelength for-

mula to find the de Broglie wavelength of the second particle:

λ′=h

p′=h

3mv

Therefore, the de Broglie wavelength of the second particle is h

3mv .

Question 8

Question

An electron is accelerated from rest through a potential difference of 100 V.

Determine the de Broglie wavelength associated with the electron after this

acceleration.

7

Solution

Step 1: Find the kinetic energy of the electron after acceleration. Given that

the potential difference is 100 V and the charge of an electron is −1.6×10−19

C, the kinetic energy can be calculated using the equation:

K.E. =q·∆V= (1.6×10−19 C) ×(100 V)

Step 2: Calculate the kinetic energy.

K.E. = 1.6×10−17 J

Step 3: Use the kinetic energy to find the de Broglie wavelength. The de

Broglie wavelength is given by:

λ=h

p=h

√2mK.E.

where his the Planck constant (6.626 ×10−34 Js), mis the mass of the electron

(9.11 ×10−31 kg), and K.E. is the kinetic energy.

Step 4: Substitute the values to find the de Broglie wavelength.

λ=6.626 ×10−34 Js

p2×9.11 ×10−31 kg ×1.6×10−17 J

Step 5: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 Js

p2×9.11 ×10−31 kg ×1.6×10−17 J

λ≈1.21 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron after ac-

celeration through a potential difference of 100 V is approximately 1.21 ×10−10

m.

Question 9

Question

An electron is accelerated through a potential difference of 1000 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the formula K=eV ,

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 2: The kinetic energy of the electron can be calculated as:

K= (1.6×10−19 C)(1000 V)

8

Step 3: Simply the expression to find the kinetic energy:

K= 1.6×10−16 J

Step 4: Using the formula for the de Broglie wavelength λ=h

p, where his

Planck’s constant (6.63 ×10−34 J s) and pis the momentum of the electron.

Step 5: The momentum of the electron is given by p=√2mK, where mis

the mass of the electron (9.11 ×10−31 kg).

Step 6: Calculate the momentum of the electron:

p=p2(9.11 ×10−31 kg)(1.6×10−16 J)

Step 7: Simplify to find the momentum:

p≈3.02 ×10−24 kg m/s

Step 8: Finally, determine the de Broglie wavelength:

λ=6.63 ×10−34 J s

3.02 ×10−24 kg m/s

Step 9: Simplify to find the de Broglie wavelength:

λ≈2.20 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 1000 V is approximately 2.20 ×10−10 m.

Question 10

Question

A beam of electrons with kinetic energy Eis directed toward a single slit of

width a. Calculate the de Broglie wavelength of the electrons after passing

through the slit.

Solution

Step 1: Calculate the momentum of the electrons using their kinetic energy.

The momentum of an electron is given by the formula p=√2mE, where mis

the mass of the electron and Eis its kinetic energy.

Step 2: Calculate the de Broglie wavelength of the electrons. The de Broglie

wavelength is given by the formula λ=h

p, where his the Planck constant.

Therefore, the de Broglie wavelength of the electrons after passing through

the slit is λ=h

√2mE .

9

Question 11

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron after acceleration.

Solution

Step 1: First, we need to calculate the kinetic energy gained by the electron

when accelerated through a potential difference of 100 V. The kinetic energy of

the electron can be calculated using the equation:

KE =qV

where KE is the kinetic energy, qis the charge of the electron, and Vis the

potential difference. Given that the charge of an electron q=−1.6×10−19 C,

and the potential difference V= 100 V, we can substitute these values into the

formula:

KE = (−1.6×10−19 C)(100 V)

KE =−1.6×10−17 J

Step 2: Next, we convert the kinetic energy into the momentum of the

electron using the equation:

KE =1

2mv2=p2

2m

where mis the mass of the electron and pis the momentum. Rearranging the

formula to solve for momentum, we get:

p=√2mKE

Given that the mass of an electron m= 9.11 ×10−31 kg, we can substitute the

values of mass and kinetic energy into the formula:

p=p2(9.11 ×10−31 kg)(1.6×10−17 J)

p≈4.88 ×10−24 kg m/s

Step 3: Finally, we can calculate the de Broglie wavelength of the electron

using the formula:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),

and pis the momentum of the electron. Substitute the values of Planck constant

and momentum into the formula to find the de Broglie wavelength:

λ=6.626 ×10−34 m2kg/s

4.88 ×10−24 kg m/s

10

λ≈1.36 ×10−10 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 100 V is approximately 1.36 ×10−10 m.

Question 12

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula KE =eV ,

where eis the elementary charge and Vis the potential difference. Step 2:

Determine the momentum of the electron using the formula p=√2mKE, where

mis the mass of the electron. Step 3: Calculate the de Broglie wavelength using

the formula λ=h

p, where his the Planck constant. Given data: V= 200 V,

e= 1.6×10−19 C, m= 9.11 ×10−31 kg, and h= 6.626 ×10−34 J·s.

Step 1: Find the kinetic energy of the electron.

KE =eV = (1.6×10−19C)(200V)=3.2×10−17J

Step 2: Determine the momentum of the electron.

p=√2mKE =p2(9.11 ×10−31kg)(3.2×10−17J)≈1.77 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength.

λ=h

p=6.626 ×10−34J·s

1.77 ×10−24 kg m/s ≈3.74 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 3.74 ×10−10 m.

Question 13

Question

An electron is accelerated through a potential difference of 150 V. Determine:

(a) The de Broglie wavelength associated with the electron. (b) The velocity of

the electron.

11

Solution

(a) Let’s first determine the de Broglie wavelength associated with the electron

using the formula:

λ=h

p

where λ= de Broglie wavelength (m), h= Planck’s constant = 6.626×10−34 Js,

p= momentum of the particle.

We know that momentum (p) can be calculated using the formula:

p=√2mE

where m= mass of the electron = 9.11 ×10−31 kg, E= kinetic energy of the

electron.

Given that the potential difference is 150 V, we can calculate the kinetic

energy using the formula:

E=qV

where q= charge of an electron = 1.6×10−19 C, V= potential difference =

150 V.

Step 1: Calculate the kinetic energy of the electron.

E=qV = (1.6×10−19 C)(150 V)

E= 2.4×10−17 J

Step 2: Calculate the momentum of the electron.

p=√2mE =p2(9.11 ×10−31 kg)(2.4×10−17 J)

p≈5.44 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength.

λ=h

p=6.626 ×10−34 Js

5.44 ×10−24 kg m/s

λ≈1.22 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron is approx-

imately 1.22 ×10−10 m.

(b) We can now determine the velocity of the electron using the de Broglie

wavelength:

v=h

mλ

where v= velocity of the electron.

Step 1: Calculate the velocity of the electron.

v=h

mλ =6.626 ×10−34 Js

(9.11 ×10−31 kg)(1.22 ×10−10 m)

v≈5.39 ×106m/s

Therefore, the velocity of the electron is approximately 5.39 ×106m/s.

12

Question 14

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula, which relates the wave-

length of a particle to its momentum. The formula is given by:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J·s),

and pis the momentum of the particle.

Step 2: First, we need to find the momentum of the electron. The kinetic

energy of the electron can be calculated using the formula:

K.E. =qV

where qis the charge of the electron (−1.6×10−19 C) and Vis the potential

difference (200 V in this case).

Step 3: Substituting the values into the formula for kinetic energy:

K.E. = (−1.6×10−19 C)(200 V) = −3.2×10−17 J

Step 4: The kinetic energy can also be expressed in terms of momentum as:

K.E. =p2

2m

where mis the mass of the electron.

Step 5: Rearranging the formula to solve for momentum, we get:

p=√2mK.E.

Step 6: Substituting the values into the formula for momentum:

p=p2×9.11 ×10−31 kg ×3.2×10−17 J≈4.44 ×10−24 kg ·m/s

Step 7: Finally, substituting the momentum into the de Broglie wavelength

formula:

λ=6.626 ×10−34 J·s

4.44 ×10−24 kg ·m/s ≈1.49 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 1.49 ×10−10 m.

13

Question 15

Question

A beam of electrons with kinetic energy 500 eV is incident on a crystal. If

the crystal diffracts the electrons at an angle of 30◦, what is the de Broglie

wavelength of the electrons?

Solution

Step 1: We first find the velocity of the electrons using the kinetic energy formula

KE =1

2mev2, where meis the mass of an electron (9.11 ×10−31 kg).

Step 2: Rearranging the kinetic energy formula gives us v=q2×500 eV×1.6×10−19 C/eV

9.11×10−31 kg .

Converting the kinetic energy to joules and solving for velocity gives v≈

2.19 ×106m/s.

Step 3: Using the definition of de Broglie wavelength λ=h

p, where his the

Planck constant (6.626 ×10−34 J s) and pis the momentum of the electron.

Step 4: The momentum of an electron can be calculated as p=mev, and

substituting the values gives p= (9.11 ×10−31 kg) ×(2.19 ×106m/s). Solving

for momentum gives p≈2.00 ×10−24 kg m/s.

Step 5: Substituting the momentum into the de Broglie wavelength formula

gives λ=6.626×10−34 J s

2.00×10−24 kg m/s . Calculating the de Broglie wavelength gives λ≈

3.31 ×10−12 m.

Therefore, the de Broglie wavelength of the electrons is approximately 3.31×

10−12 m.

Question 16

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength of the electron in meters.

Solution

Step 1: Determine the kinetic energy of the electron using the formula K=eV ,

where eis the elementary charge and Vis the potential difference. Step 2:

Use the kinetic energy of the electron to find its momentum using the formula

p=√2mK, where mis the mass of the electron and Kis the kinetic energy.

Step 3: Calculate the de Broglie wavelength using the formula λ=h

p, where h

is the Planck constant and pis the momentum of the electron. Step 4: Convert

the de Broglie wavelength to meters.

Now let’s proceed with the calculations.

Step 1: Determine the kinetic energy of the electron. Given: Potential

difference, V= 100 V Elementary charge, e= 1.6×10−19 C

14

Using K=eV :K= (1.6×10−19 C)(100 V) K= 1.6×10−17 J

Step 2: Calculate the momentum of the electron. Given: Mass of electron,

m= 9.109 ×10−31 kg

Using p=√2mK:p=p2(9.109 ×10−31 kg)(1.6×10−17 J) p≈4.740 ×

10−24 kg m/s

Step 3: Calculate the de Broglie wavelength. Given: Planck constant,

h= 6.626 ×10−34 m2kg/s

Using λ=h

p:λ=6.626×10−34 m2kg/s

4.740×10−24 kg m/s λ≈1.399 ×10−10 m

Step 4: Convert the de Broglie wavelength to meters. The de Broglie

wavelength of the electron is approximately 1.399 ×10−10 meters.

Question 17

Question

An electron is accelerated through a potential difference of 100 V. Calculate

the de Broglie wavelength of the electron after being accelerated. (Mass of an

electron = 9.11 ×10−31 kg, charge of an electron = 1.60 ×10−19 C, Planck’s

constant = 6.63 ×10−34 J·s)

Solution

Step 1: Calculate the kinetic energy of the electron. The kinetic energy of the

electron can be calculated using the formula:

KE =qV

where: KE = kinetic energy (J), q= charge of the electron (C), V= potential

difference (V).

Substitute the given values:

KE = (1.60 ×10−19 C)(100 V) = 1.60 ×10−17 J

Step 2: Calculate the velocity of the electron. The kinetic energy of the

electron can also be expressed in terms of its velocity vusing the formula:

KE =1

2mv2

where: m= mass of the electron (kg), v= velocity of the electron (m/s).

Equating the two expressions for kinetic energy:

1

2mv2= 1.60 ×10−17

Solving for v:

v=r2×1.60 ×10−17

m

15

v=r2×1.60 ×10−17

9.11 ×10−31

v=r3.20 ×10−17

9.11 ×10−31

v≈√3.514 ×107

v≈5.92 ×103m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength λof an electron is given by the formula:

λ=h

p

where: h= Planck’s constant (J·s), p= momentum of the electron (kg·m/s).

The momentum pof the electron is given by:

p=mv

p= (9.11 ×10−31 kg)(5.92 ×103m/s)

p≈5.40 ×10−27 kg ·m/s

Substitute the values into the de Broglie wavelength formula:

λ=6.63 ×10−34 J·s

5.40 ×10−27 kg·m/s

λ≈6.63 ×10−34

5.40 ×107

λ≈1.23 ×10−10 m

Therefore, the de Broglie wavelength of the electron after being accelerated

through a potential difference of 100 V is approximately 1.23 ×10−10 m.

Question 18

Question

An electron is accelerated through a potential difference of 500 V. What is the

de Broglie wavelength associated with this electron?

16

Solution

Step 1: Use the formula for de Broglie wavelength:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),

and pis the momentum of the particle.

Step 2: First, find the kinetic energy of the electron using the formula:

kinetic energy (KE) = charge ×potential difference

Given that the charge of an electron e= 1.6×10−19 C:

KE = e×potential difference

Step 3: Calculate the velocity of the electron using the formula for kinetic

energy:

KE = 1

2mv2

where mis the mass of the electron (9.11 ×10−31 kg) and vis the velocity of

the electron.

Step 4: Calculate the momentum pof the electron using its mass and veloc-

ity:

p=m×v

Step 5: Substitute the values of Planck’s constant hand momentum pinto

the de Broglie wavelength equation to find the de Broglie wavelength λ.

Question 19

Question

An electron is accelerated through a potential difference of 150 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula for particles:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J·s),

and pis the momentum of the particle.

Step 2: To find the momentum, we can use the formula for the momentum

of a particle:

p=√2mE

17

where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy

of the electron.

Step 3: The kinetic energy of the electron can be found using the formula:

E=qV

where qis the charge of the electron (−1.6×10−19 C) and Vis the potential

difference it is accelerated through (150 V).

Step 4: Substituting the kinetic energy into the momentum formula gives:

p=p2m(qV )

Step 5: Now, substituting the momentum into the de Broglie wavelength

formula gives:

λ=h

p2m(qV )

Step 6: Plugging in the values (h= 6.626 ×10−34 J·s, m= 9.11 ×10−31 kg,

q=−1.6×10−19 C, and V= 150 V) and calculating gives:

λ=6.626 ×10−34 J·s

p2(9.11 ×10−31 kg)(−1.6×10−19 C)(150 V)

Step 7: Solving for λgives the de Broglie wavelength associated with the

electron accelerated through the potential difference of 150 V.

Question 20

Question

An electron is accelerated through a potential difference of 120 V. Determine

the de Broglie wavelength of the electron after passing through the potential

difference.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

Given that the potential difference is 120 V, we can calculate the kinetic energy

using the formula K.E. =qV , where qis the charge of the electron (1.6 ×10−19

C) and Vis the potential difference.

Plugging in the values, we get:

K.E. = (1.6×10−19 C) ×(120 V) = 1.92 ×10−17 J

Step 2: Use the kinetic energy to find the momentum of the electron. The

kinetic energy of the electron can be related to its momentum using the formula

K.E. =1

2mv2=p2

2m, where mis the mass of the electron and vis its velocity.

18

Therefore, we get:

p2= 2m×K.E.

p=√2m×K.E.

Substitute the values of m(mass of electron = 9.11 ×10−31 kg) and K.E.:

p=p2×9.11 ×10−31 kg ×1.92 ×10−17 J=3.12 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength (λ) is related to the momentum of the electron by the formula λ=h

p,

where his the Planck constant (6.626 ×10−34 J s).

Substitute the values to find the de Broglie wavelength:

λ=6.626 ×10−34 J s

3.12 ×10−24 kg m/s = 2.12 ×10−10 m

Therefore, the de Broglie wavelength of the electron after passing through

the potential difference is 2.12 ×10−10 meters.

Question 21

Question

An electron is accelerated through a potential difference of 250 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula Ek=qV ,

where qis the electron charge and Vis the potential difference.

Given: V= 250 V (potential difference)

Elementary charge, e= 1.6×10−19 C

Ek=eV = (1.6×10−19 C)(250 V)

Ek= 4 ×10−17 J

Step 2: Use the formula for kinetic energy to find the momentum of the

electron.

Ek=p2

2m(where pis momentum and mis mass)

p=p2mEk

Mass of electron, m= 9.11 ×10−31 kg

p=p2(9.11 ×10−31 kg)(4 ×10−17 J)

p≈6.09 ×10−24 kg m/s

19

Step 3: Calculate the de Broglie wavelength using the formula λ=h

p,

where his the Planck constant.

h= 6.63 ×10−34 J s

λ=6.63 ×10−34 J s

6.09 ×10−24 kg m/s

λ≈1.09 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 250 V is approximately 1.09 ×10−10 m.

Question 22

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength associated with this electron.

Solution

Step 1: We can use the de Broglie wavelength formula to find the wavelength

associated with the electron:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J

·s), and pis the momentum of the particle.

Step 2: First, we need to find the momentum of the electron. The kinetic

energy of the electron can be calculated using the potential difference it was

accelerated through:

K.E. =e·V

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 3: Substitute the given values into the equation to find the kinetic

energy of the electron:

K.E. = (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 4: Next, we use the relationship between kinetic energy and momentum

to find the momentum of the electron:

K.E. =p2

2m

where mis the mass of the electron (9.11 ×10−31 kg).

Step 5: Solve for the momentum p:

p=√2m·K.E. =p2×9.11 ×10−31 kg ×1.6×10−17 J≈2.20 ×10−24 kg m/s

20

Step 6: Finally, substitute the momentum into the de Broglie wavelength

formula to find the wavelength:

λ=6.626 ×10−34 J s

2.20 ×10−24 kg m/s ≈3.01 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 100 V is approximately 3.01 ×10−10 m.

Question 23

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference

given. Step 2: Use the kinetic energy to find the velocity of the electron. Step

3: Use the velocity to calculate the de Broglie wavelength of the electron.

Step 1: Calculate the kinetic energy of the electron. Using the formula for

kinetic energy, K.E. =qV , where qis the charge of an electron (1.6×10−19 C)

and Vis the potential difference (200 V):

K.E. = (1.6×10−19 C)(200 V) = 3.2×10−17 J

Step 2: Find the velocity of the electron. Since the kinetic energy is equal

to 1

2mv2, where mis the mass of the electron (9.11 ×10−31 kg), we can solve

for the velocity v:

3.2×10−17 J = 1

2(9.11 ×10−31 kg)v2

v=s2×3.2×10−17 J

9.11 ×10−31 kg ≈6.09 ×106m/s

Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie

wavelength is given by λ=h

p, where his the Planck constant (6.626 ×10−34

J·s) and pis the momentum of the electron (m·v):

λ=6.626 ×10−34 J·s

9.11 ×10−31 kg ·6.09 ×106m/s ≈1.36 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately 1.36 ×10−10 m.

21

Question 24

Question

A proton is moving with a velocity of 2.5×106m/s. Calculate the de Broglie

wavelength associated with the proton’s motion.

Solution

Step 1: Recall the de Broglie wavelength formula:

λ=h

p

where: - λis the de Broglie wavelength, - his the Planck constant (6.626×10−34

J·s), and - pis the momentum of the particle.

Step 2: Calculate the momentum of the proton using the equation p=mv,

where mis the mass of the proton and vis its velocity.

p= (1.67 ×10−27 kg) ×(2.5×106m/s)

Step 3: Calculate the momentum of the proton.

p= 4.175 ×10−21 kg m/s

Step 4: Substitute the momentum into the de Broglie wavelength formula

to get the de Broglie wavelength for the proton.

λ=6.626 ×10−34 J·s

4.175 ×10−21 kg m/s

Step 5: Calculate the de Broglie wavelength.

λ≈1.587 ×10−13 m

Therefore, the de Broglie wavelength associated with the proton’s motion is

approximately 1.587 ×10−13 m.

Question 25

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength associated with this electron.

22

Solution

Step 1: Determine the kinetic energy of the electron using the potential differ-

ence. Step 2: Apply the de Broglie wavelength formula to find the wavelength

associated with the electron.

Step 1: The kinetic energy of the electron can be calculated as

K.E. =qV

where qis the charge of the electron and Vis the potential difference. Given that

the potential difference is 200 V and the charge of an electron is −1.6×10−19

C, we have

K.E. = (−1.6×10−19 C) ×(200 V) = −3.2×10−17 J

Step 2: The de Broglie wavelength of the electron is given by

λ=h

p

where his the Planck’s constant (6.626×10−34 m2kg/s) and pis the momentum

of the electron. The momentum of the electron can be calculated using the

kinetic energy:

p=√2mK.E.

where mis the mass of the electron. Substitute the values in to find the de

Broglie wavelength associated with the electron.

Question 26

Question

Two particles, one with mass m1and another with mass m2, are moving with

the same velocity. Calculate the ratio of their de Broglie wavelengths, λ1

λ2.

Solution

1. The de Broglie wavelength of a particle is given by:

λ=h

p

where his the Planck constant and pis the momentum of the particle.

2. The momentum pof a particle is given by:

p=mv

where mis the mass of the particle and vis its velocity.

23

3. Therefore, the de Broglie wavelength λof a particle with mass mand

velocity vcan be written as:

λ=h

mv

4. Let’s calculate the de Broglie wavelengths λ1and λ2for particles with

masses m1and m2moving at the same velocity.

For particle 1 with mass m1:

λ1=h

m1v

For particle 2 with mass m2:

λ2=h

m2v

5. Now, we’ll find the ratio λ1

λ2:

λ1

λ2

=

h

m1v

h

m2v

=m2

m1

6. Therefore, the ratio of the de Broglie wavelengths of the two particles is

m2

m1

.

Question 27

Question

An electron is accelerated through a potential difference of 100 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Use the formula relating the de Broglie wavelength (λ) with the mo-

mentum (p) of a particle:

λ=h

p

where his the Planck constant (6.63 ×10−34 m2kg/s).

Step 2: Find the momentum of the electron using its kinetic energy: The

kinetic energy of an electron accelerated through a potential difference is given

by:

K.E. =eV

24

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Given that the potential difference (V) is 100 V, the kinetic energy of the

electron is:

K.E. =e×100 = 1.6×10−17 J

Step 3: Use the relationship between kinetic energy and momentum to find

the momentum p:

K.E. =p2

2m

where mis the mass of the electron (9.11 ×10−31 kg).

Substitute the values to find the momentum:

1.6×10−17 =p2

2×9.11 ×10−31

p=p2×9.11 ×10−31 ×1.6×10−17

Step 4: Calculate the de Broglie wavelength using the momentum obtained:

Now substitute the momentum into the formula for de Broglie wavelength:

λ=6.63 ×10−34

√2×9.11 ×10−31 ×1.6×10−17

λ≈6.63 ×10−34

2.14 ×10−24 ≈3.1×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 100 V is approximately 3.1×10−10 m.

Question 28

Question

An electron is accelerated through a potential difference of 200 V. Calculate the

de Broglie wavelength of the electron after acceleration. (Take the charge of an

electron, e, to be 1.6×10−19 C.)

Solution

Step 1: Calculate the kinetic energy of the electron using the potential difference.

The kinetic energy gained by the electron is equal to the work done on it by the

potential difference:

Kinetic energy (KE) = Charge ×Potential difference

KE = e×V

KE = 1.6×10−19 C×200 V

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KE = 3.2×10−17 J

Step 2: Use the kinetic energy of the electron to find its momentum. The

momentum of the electron is given by:

Momentum = p2m×Kinetic energy

where mis the mass of the electron. The mass of an electron is 9.11 ×10−31 kg.

Momentum = p2×9.11 ×10−31 kg ×3.2×10−17 J

Momentum ≈2.19 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength using the momentum of the

electron. The de Broglie wavelength is given by:

λ=h

p

where his the Planck constant, 6.63 ×10−34 J s.

λ=6.63 ×10−34 J s

2.19 ×10−24 kg m/s

λ≈3.03 ×10−11 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 200 V is approximately 3.03 ×10−11 m.

Question 29

Question

A particle of mass m= 5.0×10−27 kg moves with a speed of 4.0×105m/s.

Calculate the de Broglie wavelength associated with this particle.

Solution

Step 1: Calculate the momentum of the particle using the formula p=mv,

where mis the mass and vis the speed.

p= (5.0×10−27 kg)(4.0×105m/s)

p= 2.0×10−21 kg m/s

Step 2: Calculate the de Broglie wavelength using the formula λ=h

p, where

his the Planck constant.

λ=6.63 ×10−34 J s

2.0×10−21 kg m/s

λ= 3.32 ×10−13 m

Therefore, the de Broglie wavelength associated with the particle is 3.32 ×

10−13 meters.

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Question 30

Question

An electron is accelerated from rest through a potential difference of 120 V.

Determine the de Broglie wavelength associated with the electron after acceler-

ation.

Solution

Step 1: We know that the kinetic energy of the electron after acceleration

through a potential difference of Vis given by the equation:

KE =eV

where eis the elementary charge and Vis the potential difference.

Step 2: Substituting the given values, we find the kinetic energy of the

electron:

KE = (1.6×10−19 C)(120 V)

KE = 1.92 ×10−17 J

Step 3: The de Broglie wavelength of a particle is given by the equation:

λ=h

p

where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J s),

and pis the momentum of the particle.

Step 4: The momentum of the electron can be calculated using its kinetic

energy:

p=p2meKE

where meis the mass of the electron (9.11 ×10−31 kg).

Step 5: Substituting the values and calculating the momentum:

p=p2(9.11 ×10−31 kg)(1.92 ×10−17 J)

p≈4.38 ×10−24 kg m/s

Step 6: Finally, we can determine the de Broglie wavelength of the electron:

λ=6.626 ×10−34 J s

4.38 ×10−24 kg m/s

λ≈1.51 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron after ac-

celeration is approximately 1.51 ×10−10 m.

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Question 31

Question

A particle of mass mmoves with a velocity v. By applying the de Broglie

hypothesis, calculate the de Broglie wavelength associated with this particle.

Solution

Step 1: According to the de Broglie hypothesis, the wavelength (λ) associated

with a particle of momentum pis given by:

λ=h

p

where his the Planck constant.

Step 2: The momentum of a particle of mass mmoving with velocity vis

given by:

p=mv

Step 3: Substitute the expression for momentum into the de Broglie wave-

length formula:

λ=h

mv

Step 4: Therefore, the de Broglie wavelength associated with a particle of

mass mmoving with velocity vis h/(mv).

Question 32

Question

An electron is accelerated from rest through a potential difference of 500 V.

Calculate the de Broglie wavelength of the electron after acceleration.

Solution

Step 1: First, we need to calculate the kinetic energy (K.E.) of the electron after

acceleration using the formula K.E. =eV , where eis the charge of an electron

and Vis the potential difference. Given e= 1.6×10−19 C and V= 500 V, we

substitute these values into the formula:

K.E. = (1.6×10−19 C) ×(500 V)

K.E. = 8 ×10−17 J

Step 2: Next, we use the kinetic energy of the electron to find its velocity

vusing the formula K.E. =1

2mv2, where mis the mass of the electron. The

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mass of an electron is 9.11 ×10−31 kg. Substituting the values into the formula,

we get:

8×10−17 J = 1

2×(9.11 ×10−31 kg) ×v2

v=s2×8×10−17 J

9.11 ×10−31 kg

v≈2.75 ×106m/s

Step 3: Finally, we can calculate the de Broglie wavelength λof the electron

using the formula λ=h

p, where his the Planck constant and pis the momentum

of the electron. The momentum pcan be calculated as p=mv. Substituting

the values into the formula, we get:

λ=h

mv =6.626 ×10−34 J s

(9.11 ×10−31 kg) ×(2.75 ×106m/s)

λ≈2.44 ×10−12 m

Therefore, the de Broglie wavelength of the electron after acceleration through

a potential difference of 500 V is approximately 2.44 ×10−12 m.

Question 33

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with the electron.

Solution

Step 1: Recall the de Broglie wavelength formula: λ=h

p, where his the Planck

constant (6.63 ×10−34 J·s) and pis the momentum of the particle.

Step 2: To find the momentum p, we can use the relation between kinetic

energy, charge, and potential difference: K.E. =qV , where qis the charge of

the electron and Vis the potential difference.

Step 3: The kinetic energy of the electron can be calculated as K.E. =eV ,

where eis the elementary charge (1.6×10−19 C) and Vis the potential difference.

Step 4: Substituting K.E. =eV into the expression for momentum p=

√2mK.E., where mis the mass of the electron, we have p=√2meV .

Step 5: Now, we have the momentum pin terms of the potential difference

V. Substituting this into the de Broglie wavelength formula λ=h

p, we can find

the de Broglie wavelength associated with the electron.

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Step 6: Substituting p=√2meV into λ=h

pgives:

λ=h

√2meV

Step 7: Plugging in the values for h,m,e, and V:

λ=6.63 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−19 C×200 V

Step 8: Now, calculate the de Broglie wavelength using the above formula.

Step 9: The de Broglie wavelength associated with the electron accelerated

through a potential difference of 200 V is approximately λ≈1.23 ×10−10 m.

Question 34

Question

An electron is accelerated through a potential difference of 100 V. Calculate the

de Broglie wavelength associated with the electron.

Solution

Step 1: Determine the kinetic energy of the electron using the potential differ-

ence. Given that the potential difference is 100 V and the charge of an electron

is −1.6×10−19 C, we can calculate the kinetic energy as:

K=qV =−1.6×10−19 C×100 V

Step 2: Calculate the de Broglie wavelength using the formula λ=h

p. Since

the kinetic energy is related to the momentum pby K=p2

2m, we can express p

in terms of Kand m:

p=√2mK

where mis the mass of an electron. We can then substitute this expression for

pinto the formula for the de Broglie wavelength:

λ=h

√2mK

Step 3: Substitute the known values and calculate the de Broglie wavelength.

Given that Planck’s constant h= 6.626 ×10−34 J s and the mass of an electron

m= 9.11 ×10−31 kg, we can substitute these values along with the calculated

kinetic energy into the formula:

λ=6.626 ×10−34 J s

p2×9.11 ×10−31 kg ×(−1.6×10−19 C×100 V)

Step 4: Simplify and calculate the de Broglie wavelength. After simplify-

ing the expression and performing the calculations, we can find the de Broglie

wavelength associated with the electron.

30

Question 35

Question

An electron is accelerated through a potential difference of 200 V. Determine

the de Broglie wavelength associated with this electron.

Solution

Step 1: Recall the de Broglie wavelength formula: The de Broglie wavelength

(λ) of a particle is given by:

λ=h

p

where his the Planck constant (6.626 ×10−34 J·s) and pis the momentum of

the particle.

Step 2: Find the momentum of the electron: The momentum of an electron

can be calculated using the equation p=√2meE, where meis the mass of an

electron (9.11 ×10−31 kg) and Eis the kinetic energy of the electron.

Step 3: Determine the kinetic energy of the electron: The kinetic energy of

the electron can be calculated using the equation E=qV , where qis the charge

of an electron (−1.6×10−19 C) and Vis the potential difference.

Step 4: Calculate the kinetic energy: Substitute the values q=−1.6×10−19

C and V= 200 V into the equation E=qV to find the kinetic energy of the

electron.

Step 5: Calculate the momentum: Substitute the calculated kinetic energy

into the equation p=√2meEto find the momentum of the electron.

Step 6: Find the de Broglie wavelength: Finally, substitute the calculated

momentum into the equation λ=h

pto determine the de Broglie wavelength

associated with the electron.

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