CHEM 105 - ELEMENTS OF GENERAL CHEMISTRY Wave-particle duality and de Broglie wavelength Question Bank Set 2

CHEM 105 - ELEMENTS OF

GENERAL CHEMISTRY -

Wave-particle duality and de Broglie

wavelength

Question Bank - Set 2

Liberty University

Question 1

Question

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

de Broglie wavelength of the electron.

Solution

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

Given that the potential difference is 200 V, we can find the kinetic energy

of the electron using the equation K.E. =qV , where qis the charge of the

electron (1.6 x 10−19 C) and Vis the potential difference (200 V). Therefore,

K.E. = (1.6×10−19 C) ×200 V. Calculating, we get: K.E. = 3.2×10−17 J.

Step 2: Use the kinetic energy to find the de Broglie wavelength of the

electron. The de Broglie wavelength of a particle can be calculated using the

equation λ=h

p, where λis the de Broglie wavelength, his the Planck constant

(6.626 x 10−34 J·s), and pis the momentum of the particle. Since the momentum

pis equal to √2mK.E. for non-relativistic particles, we can substitute this

into the equation for the de Broglie wavelength. Substitute the known values:

λ=6.626 ×10−34 J·s

p2×9.11 ×10−31 kg ×3.2×10−17 J.

Calculating, we get: λ≈3.27 ×10−10 m.

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

potential difference of 200 V is approximately 3.27 ×10−10 m.

Question 2

Question

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

Broglie wavelength associated with the electron.

Solution

Step 1: Recall the de Broglie wavelength equation:

λ=h

where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34

J·s), and pis the momentum of the electron.

Step 2: To find the momentum of the electron, we can use the kinetic energy

gained while being accelerated through the potential difference:

KE =qV

where KE is the kinetic energy gained, qis the charge of the electron (−1.6×

10−19 C), and Vis the potential difference.

Step 3: The kinetic energy can also be expressed in terms of momentum:

KE =p2

where mis the mass of the electron.

Step 4: Equating the expressions for kinetic energy, we get:

2m=qV

Step 5: Solving for momentum p, we get:

p=p2m·qV

Step 6: Substituting the values of m,q, and Vinto the equation, we get:

p=p2×9.11 ×10−31 ×1.6×10−19 ×100

Step 7: Calculate the value of momentum p.

Step 8: Once you have calculated the momentum p, substitute it back into

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

Question 3

Question

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

Broglie wavelength of the electron.

Solution

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

λ=h

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

J·s), and pis the momentum of the electron.

Step 2: The momentum of an electron can be calculated using its kinetic

energy:

KE =eV

where KE is the kinetic energy, eis the elementary charge (1.602 ×10−19 C),

and Vis the potential difference that the electron was accelerated through.

Step 3: Substituting the values into the kinetic energy formula, we find:

KE = (1.602 ×10−19 C)(100 V) = 1.602 ×10−17 J

Step 4: Since we know that kinetic energy is given by KE =p2

2m, we can

solve for p:

p=√2m·KE

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

Step 5: Substituting the values, we find:

p=p2·(9.11 ×10−31 kg) ·(1.602 ×10−17 J) ≈4.288 ×10−23 kg ·m/s

Step 6: Finally, we can calculate the de Broglie wavelength using the mo-

mentum:

λ=h

p=6.626 ×10−34 J·s

4.288 ×10−23 kg ·m/s ≈1.543 ×10−10 m

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

potential difference of 100 V is approximately 1.543 ×10−10 m.

Question 4

Question

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

de Broglie wavelength associated with this electron.

Solution

Step 1: Given that the electron is accelerated through a potential difference

of 120 V, we can find the kinetic energy of the electron using the equation

K.E. =e·V, where eis the elementary charge and Vis the potential difference.

Using e= 1.6×10−19 coulombs, we have:

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

Step 2: The de Broglie wavelength (λ) of the electron is given by the equa-

tion:

λ=h

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

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

p=√2·m·K.E.

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

Step 3: Substituting the expressions for momentum and Planck constant

into the de Broglie wavelength equation gives:

λ=h

√2·m·K.E.

Step 4: Now, substituting the given values into the expression for λgives:

λ=6.63 ×10−34J·s

p2·(9.11 ×10−31kg)·(1.92 ×10−17J)

Step 5: Calculating the de Broglie wavelength gives:

λ=6.63 ×10−34J·s

2.464 ×10−24kg ·m2/s2= 2.69 ×10−10m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 120 V is 2.69 ×10−10 meters.

Question 5

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: Calculate the kinetic energy of the electron using the formula KE =qV

where qis the charge of the electron and Vis the potential difference. Step 2:

Find the velocity of the electron using the formula KE =1

2mv2where mis the

mass of the electron. Step 3: Use the de Broglie wavelength formula λ=h

mv to

calculate the de Broglie wavelength of the electron.

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

difference, V= 100 V Charge of electron, q= 1.6×10−19 C

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

Step 2: Find the velocity of the electron. Mass of an electron, m= 9.11 ×

10−31 kg Using the kinetic energy formula: KE =1

2mv2

v=r2KE

m=s2(1.6×10−17 J)

9.11 ×10−31 kg ≈5.93 ×106m/s

Step 3: Calculate the de Broglie wavelength of the electron. Planck’s con-

stant, h= 6.626 ×10−34 J·s Using the de Broglie wavelength formula: λ=h

λ=6.626 ×10−34 J·s

(9.11 ×10−31 kg)(5.93 ×106m/s) ≈1.21 ×10−10 m

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

a potential difference of 100 V is approximately 1.21 ×10−10 m.

Question 6

Question

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

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

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

kg.)

Solution

1. First, find the kinetic energy of the electron after acceleration using the

formula:

Kinetic energy (KE) = qV

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

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

2. Next, find the speed of the electron using the formula for kinetic energy:

KE = 1

2mv2

where mis the mass of the electron and vis the speed of the electron.

−1.6×10−17 =1

2(9.11 ×10−31)v2

v=r−2×(−1.6×10−17)

9.11 ×10−31 = 1.88 ×106m/s

3. Finally, calculate the de Broglie wavelength of the electron using the

formula:

λ=h

where λis the de Broglie wavelength, his the Planck constant, mis the mass

of the electron, and vis the speed of the electron.

λ=6.63 ×10−34

(9.11 ×10−31)(1.88 ×106)= 3.44 ×10−10 m

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

the potential difference of 100 V is 3.44 ×10−10 m.

Question 7

Question

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

de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the speed of the electron using the kinetic energy acquired

from the potential difference. The kinetic energy of the electron is given by

K=eV , where eis the elementary charge and Vis the potential difference.

Given V= 100 V and e= 1.6×10−19 C, we have:

K=eV = (1.6×10−19 C)(100 V)

Step 2: Calculate the speed of the electron using its kinetic energy. The

kinetic energy of the electron can also be expressed in terms of its speed v:

K=1

2mv2

Solving for v:

v=r2K

Given the mass of an electron m= 9.11 ×10−31 kg, we can now calculate v.

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

is given by:

λ=h

p=h

where his the Planck constant. Substituting the calculated values for h,m,

and vwill give us the de Broglie wavelength associated with the electron.

Step 4: Substitute the known values and calculate the de Broglie wavelength.

λ=6.626 ×10−34 Js

(9.11 ×10−31 kg)(v)

Now with the calculated speed v, substitute into the formula above to find

the de Broglie wavelength λ.

Question 8

Question

A particle of mass mand velocity vis incident on a double slit. If the de Broglie

wavelength associated with the particle is λ, determine the expression for the

angle θat which the first minimum is observed on the screen, in terms of λ, the

distance between the slits d, and the distance between the slits and the screen

Solution

Step 1: Calculate the path difference between the waves from the two slits. The

path difference ∆xbetween the waves from the two slits to the first minimum

is given by:

∆x=dsin θ

Step 2: Use the condition for destructive interference. For the first minimum,

the path difference ∆xshould be equal to half the wavelength, λ/2. Therefore,

we have:

dsin θ=λ

Step 3: Use small angle approximation to simplify the expression. For small

angles (θ), sin θ≈θ. Therefore, we can rewrite the expression as:

dθ =λ

Step 4: Determine the angle θin terms of λ,d, and D. In a double-slit

experiment, the distance between the slits and the screen is D. Using basic

trigonometry, we have:

tan θ=D

Step 5: Substitute the value of dθ in terms of λ. Substitute dθ =λ

2into the

tangent expression:

dλ

2d=D

2=D

λ= 2D

Therefore, the de Broglie wavelength is equal to 2D.

Question 9

Question

A photon with an energy of 3.0 eV is incident on a metal surface. If the work

function of the metal is 1.5 eV, calculate the de Broglie wavelength of the ejected

photoelectron.

Solution

Step 1: Convert the given photon energy and work function from electron volts

to joules.

Photon energy (E) = 3.0 eV ×1.6×10−19 J/eV = 4.8×10−19 J

Work function = 1.5 eV ×1.6×10−19 J/eV = 2.4×10−19 J

Step 2: Calculate the kinetic energy of the ejected photoelectron using the

conservation of energy.

Kinetic energy = Photon energy−Work function = 4.8×10−19 J−2.4×10−19 J = 2.4×10−19 J

Step 3: Use the formula for the kinetic energy of an electron in terms of its

de Broglie wavelength

KE =1

2mv2=p2

2m=h2

λ2·2m

Where: KE = Kinetic energy of the electron, p= Momentum of the electron,

m= Mass of the electron, and λ= Wavelength of the electron.

Step 4: Solve for the de Broglie wavelength λof the ejected photoelectron.

λ=h

√2mKE

λ=6.63 ×10−34 J s

p2×9.11 ×10−31 kg ×2.4×10−19 J

λ=6.63 ×10−34 J s

p2×9.11 ×10−31 kg ×2.4×10−19 J= 3.3×10−10 m

Therefore, the de Broglie wavelength of the ejected photoelectron is 3.3×

10−10 m.

Question 10

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:

λ=h

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

the electron.

Step 2: To find the momentum of the electron, we can use the equation for

the kinetic energy:

K=1

2mv2

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

Step 3: The kinetic energy gained by the electron through the potential

difference can be calculated as:

K=qV

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

difference (100 V).

Step 4: Substituting the values, we have:

K= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 5: Since all the kinetic energy is converted into the electron’s kinetic

energy, we have K=1

2mv2.

Step 6: Solving for velocity v:

v=r2K

m=s2×1.6×10−17 J

9.11 ×10−31 kg

Step 7: Calculate the velocity vto find the momentum:

p=mv

Step 8: Finally, substitute the momentum pinto the de Broglie wavelength

formula:

λ=h

Question 11

Question

An electron is accelerated through a potential difference of 100 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 that the potential difference is 100 V, we can find the kinetic energy using

the equation:

K.E. =q·V

where qis the charge of an electron and Vis the potential difference. The charge

of an electron, q, is −1.6×10−19 C.

Thus,

K.E. =−1.6×10−19 ×100

K.E. =−1.6×10−17 J

Step 2: Use the kinetic energy to calculate the de Broglie wavelength of the

electron. The de Broglie wavelength of a particle can be calculated using the

equation:

λ=h

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

momentum of an electron can be written as:

p=√2mK.E.

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

kg. Substitute the values of h,m, and K.E. into the equation to find the de

Broglie wavelength.

Step 3: Calculate the de Broglie wavelength. Substitute h= 6.63 ×10−34

J.s, m= 9.11 ×10−31 kg, and K.E. =−1.6×10−17 J into the equation for p.

p=p2×9.11 ×10−31 × −1.6×10−17

p≈2.52 ×10−24 kg.m/s

Now, substitute h= 6.63 ×10−34 J.s and p= 2.52 ×10−24 kg.m/s into the

de Broglie wavelength equation:

λ=6.63 ×10−34

2.52 ×10−24

λ≈2.63 ×10−10 m

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

imately 2.63 ×10−10 m.

Question 12

Question

A beam of electrons with kinetic energy 200 eV is incident on a crystal with

spacing between lattice planes of 0.2 nm. Calculate the de Broglie wavelength

of the electrons and determine if they exhibit wave-like behavior.

Solution

Step 1: Calculate the de Broglie wavelength using the formula:

λ=h

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

Step 2: Calculate the momentum of the electrons using the formula:

p=√2mE

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

Step 3: Substitute the values into the formulas.

Step 4: Determine whether the de Broglie wavelength is comparable to the

spacing between lattice planes in the crystal.

Step 5: Compare the de Broglie wavelength to the spacing between lattice

planes to determine if the electrons exhibit wave-like behavior.

Question 13

Question

An electron is accelerated through a voltage of 450V. Calculate the de Broglie

wavelength associated with this electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula K.E. =eV ,

where eis the elementary charge and Vis the voltage. Step 2: The de Broglie

wavelength of the electron is given by λ=h

p, where his the Planck constant

and pis the momentum of the electron. Step 3: Relate the kinetic energy and

momentum of the electron using the formula p=√2mK.E., where mis the

mass of the electron. Step 4: Substitute the value of the momentum into the

de Broglie wavelength formula to find the wavelength.

Step 1: Calculate the kinetic energy of the electron. Given: V= 450 V,

e= 1.6×10−19 C

K.E. =eV = (1.6×10−19 C)(450 V) = 7.2×10−17J

Step 2: Calculate the de Broglie wavelength of the electron. Given: h=

6.626 ×10−34 J s

λ=h

Step 3: Find the momentum of the electron. Given: m= 9.11 ×10−31 kg

p=√2mK.E. =p2(9.11 ×10−31 kg)(7.2×10−17 J) = 4.75 ×10−24 kg m/s

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

λ=6.626 ×10−34 J s

4.75 ×10−24 kg m/s = 1.40 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a voltage of 450V is 1.40 ×10−10 m.

Question 14

Question

An electron with a kinetic energy of 200 eV is incident on a single slit of width

0.1 mm. Calculate the de Broglie wavelength of the electron.

Solution

Step 1: Calculate the speed of the electron using its kinetic energy. The kinetic

energy of the electron (K.E.) can be converted to Joules by multiplying by the

electron charge, e, and then converting eV to Joules:

K.E. = 200 eV ×1.6×10−19 J/eV

K.E. = 3.2×10−17 J

The kinetic energy of the electron can be equated to its kinetic energy in

terms of its speed v:

K.E. =1

2mv2

v=r2K.E.

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

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

wavelength of an object with mass mand velocity vis given by:

λ=h

p=h

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

Now, we can substitute the values we have calculated into the formula:

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×v

Step 3: Substitute the speed into the equation. Substitute the speed we

calculated in Step 1 into the formula:

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×q2K.E.

Step 4: Calculate the de Broglie wavelength. Now, substitute the values of

K.E.,m, and hinto the equation and calculate the de Broglie wavelength of

the electron. Remember to convert the result to meters.

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×q2×3.2×10−17 J

9.11×10−31 kg

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×√2×3.2×10−17

Calculating the final result gives:

λ≈Answer in meters

Question 15

Question

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

de Broglie wavelength of the electron after being accelerated.

Solution

To calculate the de Broglie wavelength of an electron accelerated through a

potential difference, we can use the de Broglie wavelength formula:

λ=h

where

p=√2mE

and Eis the kinetic energy of the electron.

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

the electron can be calculated using the equation:

E=eV

where eis the elementary charge and Vis the potential difference. Given that

e= 1.6×10−19 C and V= 100 V, we can substitute these values into the

equation:

E= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 2: Calculate the momentum of the electron. Substitute the calculated

kinetic energy into the momentum formula:

p=√2mE

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

an electron, m, is 9.11 ×10−31 kg.

p=p2(9.11 ×10−31 kg)(1.6×10−17 J)

p≈4.23 ×10−23 kg m/s

Step 3: Calculate the de Broglie wavelength. Now that we have the mo-

mentum of the electron, we can substitute it into the de Broglie wavelength

formula:

λ=h

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

λ=6.626 ×10−34 m2kg/s

4.23 ×10−23 kg m/s

λ≈1.56 ×10−10 m

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

through a potential difference of 100 V is approximately 1.56 ×10−10 m.

Question 16

Question

A particle of mass mand velocity vis moving in one dimension. Suppose the

particle is described as a wave with wavelength λ. Show that the de Broglie

wavelength of this particle is given by λ=h

mv , where his Planck’s constant.

Solution

Step 1: Recall that the de Broglie wavelength λis given by the equation λ=h

where pis the momentum of the particle.

Step 2: The momentum pof a particle is defined as p=mv, where mis the

mass of the particle and vis its velocity.

Step 3: Substituting the expression for momentum into the equation for de

Broglie wavelength, we have λ=h

mv .

Step 4: Hence, the de Broglie wavelength of the particle is given by λ=h

mv .

Question 17

Question

A particle with mass mand velocity vmoves in one dimension. Find the de

Broglie wavelength associated with this particle.

Solution

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

of a particle is given by:

λ=h

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

particle.

Step 2: Find the momentum of the particle: The momentum pof a particle

is given by:

p=m·v

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

Step 3: Substitute the momentum into the de Broglie wavelength formula:

Substitute p=m·vinto the de Broglie wavelength formula λ=h

λ=h

m·v

Step 4: Calculate the de Broglie wavelength: Substitute the values of h,m,

and vinto the formula:

λ=6.63 ×10−34

m·v

Step 5: Final answer: The de Broglie wavelength associated with a particle

of mass mand velocity vis 6.63×10−34

m·v.

Question 18

Question

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

the de Broglie wavelength associated with the electron.

Solution

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

eis the elementary charge and Vis the potential difference.

Step 1: K=eV = (1.6×10−19 C)(150 V) = 2.4×10−17 J

Step 2: Calculate the velocity of the electron using the formula K=1

2mv2,

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

Step 2: v=r2K

m=s2(2.4×10−17 J)

9.11 ×10−31 kg ≈1.23 ×106m/s

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

mv ,

where his the Planck constant, mis the mass of the electron, and vis the

velocity.

Step 3: λ=h

mv =6.63 ×10−34 J·s

(9.11 ×10−31 kg)(1.23 ×106m/s) ≈5.33 ×10−11 m

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

imately 5.33 ×10−11 meters.

Question 19

Question

An electron (mass 9.11 ×10−31 kg) is accelerated through a potential difference

of 120 V. Calculate the de Broglie wavelength of the electron after acceleration.

Solution

Step 1: Calculate the kinetic energy of the electron. Step 2: Use the de Broglie

wavelength formula to find the wavelength.

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

formula:

KE =qV

where q is the charge of the electron (1.6×10−19 C) and V is the potential

difference (120 V).

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

Step 2: The de Broglie wavelength of the electron can be calculated using

the formula:

λ=h

where - λis the de Broglie wavelength, - his the Planck constant (6.63×10−34 J·

s), - p is the momentum of the electron (p=√2mKE).

Substitute the given values:

p=p2·(9.11 ×10−31 kg)·(1.92 ×10−17 J)

p≈6.25 ×10−24 kg ·m/s

Now, calculate the de Broglie wavelength:

λ=6.63 ×10−34 J·s

6.25 ×10−24 kg ·m/s

λ≈1.06 ×10−10 m

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

approximately 1.06 ×10−10 m.

Question 20

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: Calculate the kinetic energy of the electron. Given that the potential

difference is 100 V, the kinetic energy of the electron can be calculated using

the formula qV =1

2mv2, where qis the charge of the electron (1.6 x 10−19 C),

Vis the potential difference (100 V), mis the mass of an electron (9.11 x 10−31

kg), and vis the velocity of the electron.

Step 1: KE =qV =1

2mv2

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

formula, we have v=q2qV

Step 2: v=r2qV

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

Broglie wavelength formula λ=h

p, where his the Planck constant (6.63 x 10−34

J·s) and pis the momentum of the electron. The momentum pof the electron

is given as mv.

Step 3: λ=h

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

the values of h,m, and vinto the de Broglie wavelength formula to find the de

Broglie wavelength of the electron.

Step 4: λ=6.63 ×10−34

(9.11 ×10−31)(p2(1.6×10−19)(100))

Question 21

Question

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

de Broglie wavelength of the electron after acceleration. Remember that the

charge of an electron is −1.6×10−19 C and its mass is 9.11 ×10−31 kg.

Solution

Step 1: Determine the kinetic energy of the electron. The kinetic energy of an

electron accelerated through a potential difference Vcan be calculated using

the formula:

K.E. =qV

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

Substitute the given values:

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

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

electron can also be expressed in terms of its velocity v:

K.E. =1

2mv2

where mis the mass of the electron.

Solve for v:

−1.6×10−17 =1

2(9.11 ×10−31)(v2)

v2=−2× −1.6×10−17

9.11 ×10−31

v2=3.2×10−17

9.11 ×10−31

v2≈3.51 ×106m/s

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

wavelength of a particle with mass mand velocity vis given by the formula:

λ=h

p=h

where his the Planck constant, pis the momentum of the particle, and λis the

de Broglie wavelength.

Substitute the known values:

λ=6.63 ×10−34 J·s

(9.11 ×10−31 kg)(3.51 ×106m/s)

Calculate the de Broglie wavelength:

λ≈6.63 ×10−34

3.2×10−24

λ≈2.07 ×10−10 m

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

a potential difference of 100 V is approximately 2.07 ×10−10 m.

Question 22

Question

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

the de Broglie wavelength of the electron.

Solution

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

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

Step 3: Apply the de Broglie wavelength formula to find the wavelength of the

electron.

Step 1: The kinetic energy Kof the electron can be calculated using the

potential difference V:

K=eV

where eis the elementary charge and Vis the potential difference. Given V=

120 V, e= 1.6×10−19 C, we have:

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

Step 2: The momentum pof the electron is related to its kinetic energy by:

p=√2mK

where mis the mass of the electron. The mass of the electron m= 9.11 ×10−31

kg. Substituting K= 1.92 ×10−17 J, we get:

p=p2(9.11 ×10−31 kg)(1.92 ×10−17 J)

p≈4.36 ×10−24 kg m/s

Step 3: The de Broglie wavelength λof the electron is given by:

λ=h

where his the Planck constant and pis the momentum of the electron. The

Planck constant h= 6.63 ×10−34 J s. Substituting p= 4.36 ×10−24 kg m/s,

we have:

λ=6.63 ×10−34 J s

4.36 ×10−24 kg m/s

λ≈1.52 ×10−10 m

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

potential difference of 120 V is approximately 1.52 ×10−10 m.

Question 23

Question

A particle with mass mand speed vmoves along the positive x-axis. If the

uncertainty in its position is ∆x, calculate the uncertainty in its momentum.

Given that the de Broglie wavelength of the particle is λ, express the uncertainty

in the momentum in terms of m,v, and λ.

Solution

Step 1: Recall the Heisenberg Uncertainty Principle, which states that the prod-

uct of the uncertainties in position and momentum is bounded by a constant:

∆x·∆p≥

ℏ

Where ℏis the reduced Planck’s constant.

Step 2: We know that the momentum pof the particle is related to its

wavelength λby de Broglie’s equation:

p=h

Step 3: To find the uncertainty in momentum, we can substitute ∆pinto

the uncertainty principle inequality:

∆x·h

λ≥

ℏ

Step 4: Rearrange the inequality to solve for ∆p:

∆p≥

ℏ

2·λ

∆x

Step 5: Now we can express the uncertainty in momentum in terms of m,v,

and λ. Recall that the momentum pis related to the mass mand velocity vof

the particle by:

p=m·v

Step 6: Substituting p=m·vinto ∆p≥ℏ

2·λ

∆x, we get:

m·∆v≥

ℏ

2·λ

∆x

Therefore, the uncertainty in the momentum of the particle can be expressed

in terms of m,v, and λas m·∆v≥ℏ

2·λ

∆x.

Question 24

Question

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

de Broglie wavelength associated with the electron after acceleration.

Solution

Step 1: First, calculate the kinetic energy of the electron using the formula

KE = qV , where qis the charge of an electron and Vis the potential difference.

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

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

the formula KE = 1

2mv2, where mis the mass of an electron.

1.6×10−17 J = 1

2(9.11 ×10−31 kg)v2

v=s2(1.6×10−17 J)

9.11 ×10−31 kg ≈6.6×106m/s

Step 3: Finally, determine the de Broglie wavelength using the de Broglie

wavelength formula λ=h

p, where his the Planck constant.

λ=h

mv =6.626 ×10−34 J·s

(9.11 ×10−31 kg)(6.6×106m/s)

λ≈1.22 ×10−10 m or 122 pm

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

celeration through a potential difference of 100 V is approximately 122 pm.

Question 25

Question

A particle of mass mis moving with velocity v. Determine the de Broglie

wavelength associated with the particle.

Solution

To find the de Broglie wavelength associated with the particle, we can use the

de Broglie wavelength formula:

λ=h

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

Js), and pis the momentum of the particle.

Step 1: Find the momentum of the particle. The momentum of a particle

is given by:

p=mv

Step 2: Calculate the de Broglie wavelength. Substitute the momentum

into the de Broglie wavelength formula:

λ=h

Therefore, the de Broglie wavelength associated with the particle is h

mv .

Question 26

Question

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

de Broglie wavelength of the electron.

Solution

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

ference. The kinetic energy (KE) of the electron can be calculated from the

potential difference (V) using the formula:

KE =qV

where qis the charge of the electron (1.6×10−19 C). Substitute V= 100 V and

q= 1.6×10−19 C:

KE = (1.6×10−19 C)(100 V)

Step 2: Calculate the kinetic energy in joules.

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

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

Broglie wavelength (λ) of the electron is calculated using the formula:

λ=h

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

the electron. Since p=√2mKE for an electron,

λ=h

√2mKE

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

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

λ=6.626 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−17 J

Step 5: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−17 J≈1.22 ×10−10 m

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

potential difference of 100 V is approximately 1.22 ×10−10 m.

Question 27

Question

A particle of mass mand charge qis moving with a velocity v. Determine its

de Broglie wavelength in terms of m,q, and v.

Solution

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

of a particle is given by:

λ=h

where λis the de Broglie wavelength, his the Planck constant, and pis the

momentum of the particle.

Step 2: Express momentum in terms of mass and velocity: The momentum

pof the particle is given by:

p=m·v

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

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

length equation: Substitute p=m·vinto the de Broglie wavelength equation:

λ=h

m·v

Step 4: Finalize the expression for the de Broglie wavelength: Therefore,

the de Broglie wavelength of the particle with mass m, charge qand velocity v

is:

λ=h

m·v

Question 28

Question

An electron with kinetic energy of 50 eV is incident on a crystal lattice. Cal-

culate the de Broglie wavelength of the electron and discuss the wave-particle

duality of electrons in this scenario.

Solution

Step 1: We first need to convert the electron’s kinetic energy from electron volts

to joules.

Given: Kinetic energy of electron, Ekin = 50 eV

We know that 1 electron volt is equal to 1.6×10−19 Joules. Converting the

electron’s kinetic energy:

Ekin = 50 eV ×1.6×10−19 J/eV = 8 ×10−18 J

Step 2: Using the de Broglie wavelength formula λ=h

p, where his the

Planck constant and pis the momentum of the electron. The momentum can

be calculated using the kinetic energy:

p=p2mEkin

Substitute m= 9.11 ×10−31 kg (mass of an electron) and Ekin = 8 ×10−18 J

into the formula to get the momentum.

Step 3: Once the momentum is calculated, we can find the de Broglie wave-

length by substituting this value into the de Broglie wavelength formula.

Step 4: Discussing the wave-particle duality of electrons, we see that in this

scenario, the electron exhibits characteristics of both waves and particles. It has

a de Broglie wavelength, indicating its wave-like behavior and diffraction when

interacting with the crystal lattice. At the same time, it also has kinetic energy

and momentum, showing its particle-like nature. This duality is a fundamental

aspect of quantum mechanics, where particles such as electrons can exhibit

wave-like behaviors under certain conditions.

Question 29

Question

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

Calculate the de Broglie wavelength of the electron after acceleration. Given

that the mass of an electron is 9.11 ×10−31 kg and the elementary charge is

1.60 ×10−19 C.

Solution

Step 1: Calculate the kinetic energy of the electron after acceleration using the

formula K=qV , where qis the charge of the electron and Vis the potential

difference. Step 2: Use the kinetic energy calculated in Step 1 to find the velocity

of the electron using the formula K=1

2mv2. Step 3: Calculate the de Broglie

wavelength using the formula λ=h

mv , where his the Planck’s constant, mis

the mass of the electron, and vis the velocity of the electron.

Step 1: Given potential difference, V= 200 V Charge of an electron,

q= 1.60 ×10−19 C

The kinetic energy of the electron after acceleration is

K=qV = (1.60 ×10−19 C)(200 V) = 3.20 ×10−17 J

Step 2: Mass of the electron, m= 9.11 ×10−31 kg

Using the kinetic energy formula:

2mv2=K

2(9.11 ×10−31 kg)v2= 3.20 ×10−17 J

v2=2(3.20 ×10−17 J)

9.11 ×10−31 kg

v=r2(3.20 ×10−17)

9.11 ×10−31 ≈3.50 ×106m/s

Step 3: Planck’s constant, h= 6.63 ×10−34 Js

Using the de Broglie wavelength formula:

λ=h

λ=6.63 ×10−34

9.11 ×10−31 ×3.50 ×106

λ≈2.05 ×10−10 m

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

approximately 2.05 ×10−10 meters.

Question 30

Question

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

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

of an electron as 1.6×10−19 C, and the mass of an electron as 9.11 ×10−31 kg.)

Solution

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

difference.

Given: Voltage (V) = 100 V Charge of an electron (q) = 1.6×10−19 C

The work done in accelerating the electron through a potential difference V

is equal to the kinetic energy acquired by the electron:

Kinetic energy (KE) = qV

KE = (1.6×10−19 C)(100 V)

KE = 1.6×10−17 J

Step 2: Calculate the velocity of the electron using the kinetic energy ac-

quired.

The kinetic energy of the electron is given by:

KE =1

2mv2

where - m is the mass of the electron, - v is the velocity of the electron.

Rearranging the formula:

v=r2KE

v=s2(1.6×10−17 J)

9.11 ×10−31 kg

Calculating the velocity:

v≈5.93 ×106m/s

Step 3: Calculate the de Broglie wavelength of the electron using its velocity.

The de Broglie wavelength of a particle is given by:

λ=h

where - h is the Planck constant (6.63 ×10−34 J s), - p is the momentum of the

particle.

The momentum of the electron is given by:

p=mv

p= (9.11 ×10−31 kg)(5.93 ×106m/s)

Calculating the momentum:

p≈5.41 ×10−24 kg m/s

Now, substitute the values into the de Broglie wavelength formula:

λ=6.63 ×10−34 J s

5.41 ×10−24 kg m/s

λ≈1.23 ×10−10 m

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

approximately 1.23 ×10−10 m.

Question 31

Question

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

de Broglie wavelength associated with this electron.

Solution

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

where eis the elementary charge and Vis the potential difference.

K= (1.6×10−19 C)(200 V) = 3.2×10−17 J

Step 2: Use the formula for the de Broglie wavelength λ=h

p, where his the

Planck constant and pis the momentum of the electron. The momentum can

be expressed in terms of kinetic energy as p=√2mK, where mis the mass of

the electron.

p=p2×(9.11 ×10−31 kg) ×3.2×10−17 J = 2.93 ×10−24 kg m/s

λ=6.626 ×10−34 J s

2.93 ×10−24 kg m/s= 2.26 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron is 2.26 ×

10−10 m.

Question 32

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.

Step 2: Use the kinetic energy to find the momentum of the electron. Step 3:

Apply the de Broglie wavelength formula to calculate the wavelength.

Step 1: Calculate the kinetic energy of the electron using E=qV where

qis the charge and Vis the potential difference. Given V= 200 V and q=

1.6×10−19 C for the charge of an electron, the kinetic energy Eis:

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

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

kinetic energy Ecan be related to the momentum pusing the formula E=p2

2m,

where mis the mass of the electron. Solving for p, we get:

p=√2mE

Substitute m= 9.11 ×10−31 kg as the mass of an electron:

p=p2(9.11 ×10−31 kg)(3.2×10−17 J) ≈1.48 ×10−24 kg m/s

Step 3: Apply the de Broglie wavelength formula to calculate the wave-

length. The de Broglie wavelength λis given by:

λ=h

where his the Planck constant (6.63 ×10−34 J s). Substitute p= 1.48 ×

10−24 kg m/s:

λ=6.63 ×10−34 J s

1.48 ×10−24 kg m/s ≈4.48 ×10−10 m

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

imately 4.48 ×10−10 m.

Question 33

Question

An electron is accelerated through a potential difference of 54.6 V. Calculate the

de Broglie wavelength of the electron after it has been accelerated. (Electron

charge: e= 1.6×10−19 C, electron mass: m= 9.11 ×10−31 kg)

Solution

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

where eis the charge of the electron and Vis the potential difference.

Given: e= 1.6×10−19 C, V= 54.6 V

K=eV = (1.6×10−19 C)(54.6 V) = 8.736 ×10−18 J

Step 2: Use the kinetic energy of the electron to find its momentum (p) using

the equation K=p2

2m, where mis the mass of the electron.

p=√2mK =p2·(9.11 ×10−31 kg) ·(8.736 ×10−18 J)

p≈2.953 ×10−24 kg m/s

Step 3: Calculate the de Broglie wavelength of the electron using the formula

λ=h

p, where his the Planck constant.

Given: h= 6.63 ×10−34 J s

λ=6.63 ×10−34 J s

2.953 ×10−24 kg m/s

λ≈2.24 ×10−10 m

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

through a potential difference of 54.6 V is approximately 2.24 ×10−10 m.

Question 34

Question

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

de Broglie wavelength associated with the electron’s motion.

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: Determine the de Broglie wavelength using the velocity found in the

previous step.

Step 1: The kinetic energy (K.E.) gained by the electron when accelerated

through a potential difference (V) is given by the equation:

K.E. =qV

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

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

V= 500 V.

Substitute the values into the formula:

K.E. = (1.6×10−19 C)(500 V) = 8.0×10−17 J

Step 2: The kinetic energy of the electron is related to its velocity (v) by

the formula:

K.E. =1

2mv2

where mis the mass of the electron.

The mass of an electron m= 9.11 ×10−31 kg.

Substitute the values into the formula and solve for the velocity:

8.0×10−17 =1

2(9.11 ×10−31)v2

v2=2(8.0×10−17)

9.11 ×10−31 = 1.75 ×106

v≈1.32 ×103m/s

Step 3: The de Broglie wavelength (λ) of a particle is related to its velocity

by the formula:

λ=h

p=h

where his the Planck constant, mis the mass of the particle, vis the velocity

of the particle.

The Planck constant h= 6.63 ×10−34 J·s.

Substitute the values into the formula:

λ=6.63 ×10−34

(9.11 ×10−31)(1.32 ×103)

λ≈6.63 ×10−34

1.20 ×10−27

λ≈5.53 ×10−7m

Therefore, the de Broglie wavelength associated with the electron’s motion

is approximately 5.53 ×10−7meters.

Question 35

Question

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

the de Broglie wavelength of the electron after acceleration. (Given: Mass of

electron, m= 9.1×10−31 kg; Charge of electron, e= 1.6×10−19 C; Planck’s

constant, h= 6.63 ×10−34 J·s.)

Solution

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

The kinetic energy of the electron can be calculated using the potential energy

gained from the potential difference:

Kinetic energy (KE) = Charge ×Potential difference

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

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

kinetic energy of the electron can also be expressed in terms of its velocity:

KE = 1

2mv2

Solving for the velocity v:

v=r2×KE

m=s2×3.2×10−17 J

9.1×10−31 kg = 2.2×106m/s

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

wavelength λof an object with momentum pis given by the equation:

λ=h

Question 2

Question

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

Broglie wavelength associated with the electron.

Solution

Step 1: Recall the de Broglie wavelength equation:

λ=h

where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34

J·s), and pis the momentum of the electron.

Step 2: To find the momentum of the electron, we can use the kinetic energy

gained while being accelerated through the potential difference:

KE =qV

where KE is the kinetic energy gained, qis the charge of the electron (−1.6×

10−19 C), and Vis the potential difference.

Step 3: The kinetic energy can also be expressed in terms of momentum:

KE =p2

where mis the mass of the electron.

Step 4: Equating the expressions for kinetic energy, we get:

2m=qV

Step 5: Solving for momentum p, we get:

p=p2m·qV

Step 6: Substituting the values of m,q, and Vinto the equation, we get:

p=p2×9.11 ×10−31 ×1.6×10−19 ×100

Step 7: Calculate the value of momentum p.

Step 8: Once you have calculated the momentum p, substitute it back into

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

Question 3

Question

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

Broglie wavelength of the electron.

Solution

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

λ=h

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

J·s), and pis the momentum of the electron.

Step 2: The momentum of an electron can be calculated using its kinetic

energy:

KE =eV

where KE is the kinetic energy, eis the elementary charge (1.602 ×10−19 C),

and Vis the potential difference that the electron was accelerated through.

Step 3: Substituting the values into the kinetic energy formula, we find:

KE = (1.602 ×10−19 C)(100 V) = 1.602 ×10−17 J

Step 4: Since we know that kinetic energy is given by KE =p2

2m, we can

solve for p:

p=√2m·KE

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

Step 5: Substituting the values, we find:

p=p2·(9.11 ×10−31 kg) ·(1.602 ×10−17 J) ≈4.288 ×10−23 kg ·m/s

Step 6: Finally, we can calculate the de Broglie wavelength using the mo-

mentum:

λ=h

p=6.626 ×10−34 J·s

4.288 ×10−23 kg ·m/s ≈1.543 ×10−10 m

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

potential difference of 100 V is approximately 1.543 ×10−10 m.

Question 4

Question

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

de Broglie wavelength associated with this electron.

Solution

Step 1: Given that the electron is accelerated through a potential difference

of 120 V, we can find the kinetic energy of the electron using the equation

K.E. =e·V, where eis the elementary charge and Vis the potential difference.

Using e= 1.6×10−19 coulombs, we have:

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

Step 2: The de Broglie wavelength (λ) of the electron is given by the equa-

tion:

λ=h

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

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

p=√2·m·K.E.

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

Step 3: Substituting the expressions for momentum and Planck constant

into the de Broglie wavelength equation gives:

λ=h

√2·m·K.E.

Step 4: Now, substituting the given values into the expression for λgives:

λ=6.63 ×10−34J·s

p2·(9.11 ×10−31kg)·(1.92 ×10−17J)

Step 5: Calculating the de Broglie wavelength gives:

λ=6.63 ×10−34J·s

2.464 ×10−24kg ·m2/s2= 2.69 ×10−10m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a potential difference of 120 V is 2.69 ×10−10 meters.

Question 5

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: Calculate the kinetic energy of the electron using the formula KE =qV

where qis the charge of the electron and Vis the potential difference. Step 2:

Find the velocity of the electron using the formula KE =1

2mv2where mis the

mass of the electron. Step 3: Use the de Broglie wavelength formula λ=h

mv to

calculate the de Broglie wavelength of the electron.

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

difference, V= 100 V Charge of electron, q= 1.6×10−19 C

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

Step 2: Find the velocity of the electron. Mass of an electron, m= 9.11 ×

10−31 kg Using the kinetic energy formula: KE =1

2mv2

v=r2KE

m=s2(1.6×10−17 J)

9.11 ×10−31 kg ≈5.93 ×106m/s

Step 3: Calculate the de Broglie wavelength of the electron. Planck’s con-

stant, h= 6.626 ×10−34 J·s Using the de Broglie wavelength formula: λ=h

λ=6.626 ×10−34 J·s

(9.11 ×10−31 kg)(5.93 ×106m/s) ≈1.21 ×10−10 m

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

a potential difference of 100 V is approximately 1.21 ×10−10 m.

Question 6

Question

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

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

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

kg.)

Solution

1. First, find the kinetic energy of the electron after acceleration using the

formula:

Kinetic energy (KE) = qV

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

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

2. Next, find the speed of the electron using the formula for kinetic energy:

KE = 1

2mv2

where mis the mass of the electron and vis the speed of the electron.

−1.6×10−17 =1

2(9.11 ×10−31)v2

v=r−2×(−1.6×10−17)

9.11 ×10−31 = 1.88 ×106m/s

3. Finally, calculate the de Broglie wavelength of the electron using the

formula:

λ=h

where λis the de Broglie wavelength, his the Planck constant, mis the mass

of the electron, and vis the speed of the electron.

λ=6.63 ×10−34

(9.11 ×10−31)(1.88 ×106)= 3.44 ×10−10 m

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

the potential difference of 100 V is 3.44 ×10−10 m.

Question 7

Question

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

de Broglie wavelength associated with the electron.

Solution

Step 1: Calculate the speed of the electron using the kinetic energy acquired

from the potential difference. The kinetic energy of the electron is given by

K=eV , where eis the elementary charge and Vis the potential difference.

Given V= 100 V and e= 1.6×10−19 C, we have:

K=eV = (1.6×10−19 C)(100 V)

Step 2: Calculate the speed of the electron using its kinetic energy. The

kinetic energy of the electron can also be expressed in terms of its speed v:

K=1

2mv2

Solving for v:

v=r2K

Given the mass of an electron m= 9.11 ×10−31 kg, we can now calculate v.

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

is given by:

λ=h

p=h

where his the Planck constant. Substituting the calculated values for h,m,

and vwill give us the de Broglie wavelength associated with the electron.

Step 4: Substitute the known values and calculate the de Broglie wavelength.

λ=6.626 ×10−34 Js

(9.11 ×10−31 kg)(v)

Now with the calculated speed v, substitute into the formula above to find

the de Broglie wavelength λ.

Question 8

Question

A particle of mass mand velocity vis incident on a double slit. If the de Broglie

wavelength associated with the particle is λ, determine the expression for the

angle θat which the first minimum is observed on the screen, in terms of λ, the

distance between the slits d, and the distance between the slits and the screen

Solution

Step 1: Calculate the path difference between the waves from the two slits. The

path difference ∆xbetween the waves from the two slits to the first minimum

is given by:

∆x=dsin θ

Step 2: Use the condition for destructive interference. For the first minimum,

the path difference ∆xshould be equal to half the wavelength, λ/2. Therefore,

we have:

dsin θ=λ

Step 3: Use small angle approximation to simplify the expression. For small

angles (θ), sin θ≈θ. Therefore, we can rewrite the expression as:

dθ =λ

Step 4: Determine the angle θin terms of λ,d, and D. In a double-slit

experiment, the distance between the slits and the screen is D. Using basic

trigonometry, we have:

tan θ=D

Step 5: Substitute the value of dθ in terms of λ. Substitute dθ =λ

2into the

tangent expression:

dλ

2d=D

2=D

λ= 2D

Therefore, the de Broglie wavelength is equal to 2D.

Question 9

Question

A photon with an energy of 3.0 eV is incident on a metal surface. If the work

function of the metal is 1.5 eV, calculate the de Broglie wavelength of the ejected

photoelectron.

Solution

Step 1: Convert the given photon energy and work function from electron volts

to joules.

Photon energy (E) = 3.0 eV ×1.6×10−19 J/eV = 4.8×10−19 J

Work function = 1.5 eV ×1.6×10−19 J/eV = 2.4×10−19 J

Step 2: Calculate the kinetic energy of the ejected photoelectron using the

conservation of energy.

Kinetic energy = Photon energy−Work function = 4.8×10−19 J−2.4×10−19 J = 2.4×10−19 J

Step 3: Use the formula for the kinetic energy of an electron in terms of its

de Broglie wavelength

KE =1

2mv2=p2

2m=h2

λ2·2m

Where: KE = Kinetic energy of the electron, p= Momentum of the electron,

m= Mass of the electron, and λ= Wavelength of the electron.

Step 4: Solve for the de Broglie wavelength λof the ejected photoelectron.

λ=h

√2mKE

λ=6.63 ×10−34 J s

p2×9.11 ×10−31 kg ×2.4×10−19 J

λ=6.63 ×10−34 J s

p2×9.11 ×10−31 kg ×2.4×10−19 J= 3.3×10−10 m

Therefore, the de Broglie wavelength of the ejected photoelectron is 3.3×

10−10 m.

Question 10

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:

λ=h

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

the electron.

Step 2: To find the momentum of the electron, we can use the equation for

the kinetic energy:

K=1

2mv2

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

Step 3: The kinetic energy gained by the electron through the potential

difference can be calculated as:

K=qV

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

difference (100 V).

Step 4: Substituting the values, we have:

K= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 5: Since all the kinetic energy is converted into the electron’s kinetic

energy, we have K=1

2mv2.

Step 6: Solving for velocity v:

v=r2K

m=s2×1.6×10−17 J

9.11 ×10−31 kg

Step 7: Calculate the velocity vto find the momentum:

p=mv

Step 8: Finally, substitute the momentum pinto the de Broglie wavelength

formula:

λ=h

Question 11

Question

An electron is accelerated through a potential difference of 100 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 that the potential difference is 100 V, we can find the kinetic energy using

the equation:

K.E. =q·V

where qis the charge of an electron and Vis the potential difference. The charge

of an electron, q, is −1.6×10−19 C.

Thus,

K.E. =−1.6×10−19 ×100

K.E. =−1.6×10−17 J

Step 2: Use the kinetic energy to calculate the de Broglie wavelength of the

electron. The de Broglie wavelength of a particle can be calculated using the

equation:

λ=h

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

momentum of an electron can be written as:

p=√2mK.E.

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

kg. Substitute the values of h,m, and K.E. into the equation to find the de

Broglie wavelength.

Step 3: Calculate the de Broglie wavelength. Substitute h= 6.63 ×10−34

J.s, m= 9.11 ×10−31 kg, and K.E. =−1.6×10−17 J into the equation for p.

p=p2×9.11 ×10−31 × −1.6×10−17

p≈2.52 ×10−24 kg.m/s

Now, substitute h= 6.63 ×10−34 J.s and p= 2.52 ×10−24 kg.m/s into the

de Broglie wavelength equation:

λ=6.63 ×10−34

2.52 ×10−24

λ≈2.63 ×10−10 m

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

imately 2.63 ×10−10 m.

Question 12

Question

A beam of electrons with kinetic energy 200 eV is incident on a crystal with

spacing between lattice planes of 0.2 nm. Calculate the de Broglie wavelength

of the electrons and determine if they exhibit wave-like behavior.

Solution

Step 1: Calculate the de Broglie wavelength using the formula:

λ=h

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

Step 2: Calculate the momentum of the electrons using the formula:

p=√2mE

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

Step 3: Substitute the values into the formulas.

Step 4: Determine whether the de Broglie wavelength is comparable to the

spacing between lattice planes in the crystal.

Step 5: Compare the de Broglie wavelength to the spacing between lattice

planes to determine if the electrons exhibit wave-like behavior.

Question 13

Question

An electron is accelerated through a voltage of 450V. Calculate the de Broglie

wavelength associated with this electron.

Solution

Step 1: Find the kinetic energy of the electron using the formula K.E. =eV ,

where eis the elementary charge and Vis the voltage. Step 2: The de Broglie

wavelength of the electron is given by λ=h

p, where his the Planck constant

and pis the momentum of the electron. Step 3: Relate the kinetic energy and

momentum of the electron using the formula p=√2mK.E., where mis the

mass of the electron. Step 4: Substitute the value of the momentum into the

de Broglie wavelength formula to find the wavelength.

Step 1: Calculate the kinetic energy of the electron. Given: V= 450 V,

e= 1.6×10−19 C

K.E. =eV = (1.6×10−19 C)(450 V) = 7.2×10−17J

Step 2: Calculate the de Broglie wavelength of the electron. Given: h=

6.626 ×10−34 J s

λ=h

Step 3: Find the momentum of the electron. Given: m= 9.11 ×10−31 kg

p=√2mK.E. =p2(9.11 ×10−31 kg)(7.2×10−17 J) = 4.75 ×10−24 kg m/s

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

λ=6.626 ×10−34 J s

4.75 ×10−24 kg m/s = 1.40 ×10−10 m

Therefore, the de Broglie wavelength associated with the electron accelerated

through a voltage of 450V is 1.40 ×10−10 m.

Question 14

Question

An electron with a kinetic energy of 200 eV is incident on a single slit of width

0.1 mm. Calculate the de Broglie wavelength of the electron.

Solution

Step 1: Calculate the speed of the electron using its kinetic energy. The kinetic

energy of the electron (K.E.) can be converted to Joules by multiplying by the

electron charge, e, and then converting eV to Joules:

K.E. = 200 eV ×1.6×10−19 J/eV

K.E. = 3.2×10−17 J

The kinetic energy of the electron can be equated to its kinetic energy in

terms of its speed v:

K.E. =1

2mv2

v=r2K.E.

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

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

wavelength of an object with mass mand velocity vis given by:

λ=h

p=h

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

Now, we can substitute the values we have calculated into the formula:

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×v

Step 3: Substitute the speed into the equation. Substitute the speed we

calculated in Step 1 into the formula:

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×q2K.E.

Step 4: Calculate the de Broglie wavelength. Now, substitute the values of

K.E.,m, and hinto the equation and calculate the de Broglie wavelength of

the electron. Remember to convert the result to meters.

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×q2×3.2×10−17 J

9.11×10−31 kg

λ=6.626 ×10−34 J s

9.11 ×10−31 kg ×√2×3.2×10−17

Calculating the final result gives:

λ≈Answer in meters

Question 15

Question

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

de Broglie wavelength of the electron after being accelerated.

Solution

To calculate the de Broglie wavelength of an electron accelerated through a

potential difference, we can use the de Broglie wavelength formula:

λ=h

where

p=√2mE

and Eis the kinetic energy of the electron.

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

the electron can be calculated using the equation:

E=eV

where eis the elementary charge and Vis the potential difference. Given that

e= 1.6×10−19 C and V= 100 V, we can substitute these values into the

equation:

E= (1.6×10−19 C)(100 V) = 1.6×10−17 J

Step 2: Calculate the momentum of the electron. Substitute the calculated

kinetic energy into the momentum formula:

p=√2mE

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

an electron, m, is 9.11 ×10−31 kg.

p=p2(9.11 ×10−31 kg)(1.6×10−17 J)

p≈4.23 ×10−23 kg m/s

Step 3: Calculate the de Broglie wavelength. Now that we have the mo-

mentum of the electron, we can substitute it into the de Broglie wavelength

formula:

λ=h

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

λ=6.626 ×10−34 m2kg/s

4.23 ×10−23 kg m/s

λ≈1.56 ×10−10 m

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

through a potential difference of 100 V is approximately 1.56 ×10−10 m.

Question 16

Question

A particle of mass mand velocity vis moving in one dimension. Suppose the

particle is described as a wave with wavelength λ. Show that the de Broglie

wavelength of this particle is given by λ=h

mv , where his Planck’s constant.

Solution

Step 1: Recall that the de Broglie wavelength λis given by the equation λ=h

where pis the momentum of the particle.

Step 2: The momentum pof a particle is defined as p=mv, where mis the

mass of the particle and vis its velocity.

Step 3: Substituting the expression for momentum into the equation for de

Broglie wavelength, we have λ=h

mv .

Step 4: Hence, the de Broglie wavelength of the particle is given by λ=h

mv .

Question 17

Question

A particle with mass mand velocity vmoves in one dimension. Find the de

Broglie wavelength associated with this particle.

Solution

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

of a particle is given by:

λ=h

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

particle.

Step 2: Find the momentum of the particle: The momentum pof a particle

is given by:

p=m·v

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

Step 3: Substitute the momentum into the de Broglie wavelength formula:

Substitute p=m·vinto the de Broglie wavelength formula λ=h

λ=h

m·v

Step 4: Calculate the de Broglie wavelength: Substitute the values of h,m,

and vinto the formula:

λ=6.63 ×10−34

m·v

Step 5: Final answer: The de Broglie wavelength associated with a particle

of mass mand velocity vis 6.63×10−34

m·v.

Question 18

Question

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

the de Broglie wavelength associated with the electron.

Solution

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

eis the elementary charge and Vis the potential difference.

Step 1: K=eV = (1.6×10−19 C)(150 V) = 2.4×10−17 J

Step 2: Calculate the velocity of the electron using the formula K=1

2mv2,

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

Step 2: v=r2K

m=s2(2.4×10−17 J)

9.11 ×10−31 kg ≈1.23 ×106m/s

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

mv ,

where his the Planck constant, mis the mass of the electron, and vis the

velocity.

Step 3: λ=h

mv =6.63 ×10−34 J·s

(9.11 ×10−31 kg)(1.23 ×106m/s) ≈5.33 ×10−11 m

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

imately 5.33 ×10−11 meters.

Question 19

Question

An electron (mass 9.11 ×10−31 kg) is accelerated through a potential difference

of 120 V. Calculate the de Broglie wavelength of the electron after acceleration.

Solution

Step 1: Calculate the kinetic energy of the electron. Step 2: Use the de Broglie

wavelength formula to find the wavelength.

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

formula:

KE =qV

where q is the charge of the electron (1.6×10−19 C) and V is the potential

difference (120 V).

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

Step 2: The de Broglie wavelength of the electron can be calculated using

the formula:

λ=h

where - λis the de Broglie wavelength, - his the Planck constant (6.63×10−34 J·

s), - p is the momentum of the electron (p=√2mKE).

Substitute the given values:

p=p2·(9.11 ×10−31 kg)·(1.92 ×10−17 J)

p≈6.25 ×10−24 kg ·m/s

Now, calculate the de Broglie wavelength:

λ=6.63 ×10−34 J·s

6.25 ×10−24 kg ·m/s

λ≈1.06 ×10−10 m

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

approximately 1.06 ×10−10 m.

Question 20

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: Calculate the kinetic energy of the electron. Given that the potential

difference is 100 V, the kinetic energy of the electron can be calculated using

the formula qV =1

2mv2, where qis the charge of the electron (1.6 x 10−19 C),

Vis the potential difference (100 V), mis the mass of an electron (9.11 x 10−31

kg), and vis the velocity of the electron.

Step 1: KE =qV =1

2mv2

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

formula, we have v=q2qV

Step 2: v=r2qV

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

Broglie wavelength formula λ=h

p, where his the Planck constant (6.63 x 10−34

J·s) and pis the momentum of the electron. The momentum pof the electron

is given as mv.

Step 3: λ=h

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

the values of h,m, and vinto the de Broglie wavelength formula to find the de

Broglie wavelength of the electron.

Step 4: λ=6.63 ×10−34

(9.11 ×10−31)(p2(1.6×10−19)(100))

Question 21

Question

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

de Broglie wavelength of the electron after acceleration. Remember that the

charge of an electron is −1.6×10−19 C and its mass is 9.11 ×10−31 kg.

Solution

Step 1: Determine the kinetic energy of the electron. The kinetic energy of an

electron accelerated through a potential difference Vcan be calculated using

the formula:

K.E. =qV

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

Substitute the given values:

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

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

electron can also be expressed in terms of its velocity v:

K.E. =1

2mv2

where mis the mass of the electron.

Solve for v:

−1.6×10−17 =1

2(9.11 ×10−31)(v2)

v2=−2× −1.6×10−17

9.11 ×10−31

v2=3.2×10−17

9.11 ×10−31

v2≈3.51 ×106m/s

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

wavelength of a particle with mass mand velocity vis given by the formula:

λ=h

p=h

where his the Planck constant, pis the momentum of the particle, and λis the

de Broglie wavelength.

Substitute the known values:

λ=6.63 ×10−34 J·s

(9.11 ×10−31 kg)(3.51 ×106m/s)

Calculate the de Broglie wavelength:

λ≈6.63 ×10−34

3.2×10−24

λ≈2.07 ×10−10 m

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

a potential difference of 100 V is approximately 2.07 ×10−10 m.

Question 22

Question

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

the de Broglie wavelength of the electron.

Solution

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

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

Step 3: Apply the de Broglie wavelength formula to find the wavelength of the

electron.

Step 1: The kinetic energy Kof the electron can be calculated using the

potential difference V:

K=eV

where eis the elementary charge and Vis the potential difference. Given V=

120 V, e= 1.6×10−19 C, we have:

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

Step 2: The momentum pof the electron is related to its kinetic energy by:

p=√2mK

where mis the mass of the electron. The mass of the electron m= 9.11 ×10−31

kg. Substituting K= 1.92 ×10−17 J, we get:

p=p2(9.11 ×10−31 kg)(1.92 ×10−17 J)

p≈4.36 ×10−24 kg m/s

Step 3: The de Broglie wavelength λof the electron is given by:

λ=h

where his the Planck constant and pis the momentum of the electron. The

Planck constant h= 6.63 ×10−34 J s. Substituting p= 4.36 ×10−24 kg m/s,

we have:

λ=6.63 ×10−34 J s

4.36 ×10−24 kg m/s

λ≈1.52 ×10−10 m

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

potential difference of 120 V is approximately 1.52 ×10−10 m.

Question 23

Question

A particle with mass mand speed vmoves along the positive x-axis. If the

uncertainty in its position is ∆x, calculate the uncertainty in its momentum.

Given that the de Broglie wavelength of the particle is λ, express the uncertainty

in the momentum in terms of m,v, and λ.

Solution

Step 1: Recall the Heisenberg Uncertainty Principle, which states that the prod-

uct of the uncertainties in position and momentum is bounded by a constant:

∆x·∆p≥

ℏ

Where ℏis the reduced Planck’s constant.

Step 2: We know that the momentum pof the particle is related to its

wavelength λby de Broglie’s equation:

p=h

Step 3: To find the uncertainty in momentum, we can substitute ∆pinto

the uncertainty principle inequality:

∆x·h

λ≥

ℏ

Step 4: Rearrange the inequality to solve for ∆p:

∆p≥

ℏ

2·λ

∆x

Step 5: Now we can express the uncertainty in momentum in terms of m,v,

and λ. Recall that the momentum pis related to the mass mand velocity vof

the particle by:

p=m·v

Step 6: Substituting p=m·vinto ∆p≥ℏ

2·λ

∆x, we get:

m·∆v≥

ℏ

2·λ

∆x

Therefore, the uncertainty in the momentum of the particle can be expressed

in terms of m,v, and λas m·∆v≥ℏ

2·λ

∆x.

Question 24

Question

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

de Broglie wavelength associated with the electron after acceleration.

Solution

Step 1: First, calculate the kinetic energy of the electron using the formula

KE = qV , where qis the charge of an electron and Vis the potential difference.

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

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

the formula KE = 1

2mv2, where mis the mass of an electron.

1.6×10−17 J = 1

2(9.11 ×10−31 kg)v2

v=s2(1.6×10−17 J)

9.11 ×10−31 kg ≈6.6×106m/s

Step 3: Finally, determine the de Broglie wavelength using the de Broglie

wavelength formula λ=h

p, where his the Planck constant.

λ=h

mv =6.626 ×10−34 J·s

(9.11 ×10−31 kg)(6.6×106m/s)

λ≈1.22 ×10−10 m or 122 pm

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

celeration through a potential difference of 100 V is approximately 122 pm.

Question 25

Question

A particle of mass mis moving with velocity v. Determine the de Broglie

wavelength associated with the particle.

Solution

To find the de Broglie wavelength associated with the particle, we can use the

de Broglie wavelength formula:

λ=h

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

Js), and pis the momentum of the particle.

Step 1: Find the momentum of the particle. The momentum of a particle

is given by:

p=mv

Step 2: Calculate the de Broglie wavelength. Substitute the momentum

into the de Broglie wavelength formula:

λ=h

Therefore, the de Broglie wavelength associated with the particle is h

mv .

Question 26

Question

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

de Broglie wavelength of the electron.

Solution

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

ference. The kinetic energy (KE) of the electron can be calculated from the

potential difference (V) using the formula:

KE =qV

where qis the charge of the electron (1.6×10−19 C). Substitute V= 100 V and

q= 1.6×10−19 C:

KE = (1.6×10−19 C)(100 V)

Step 2: Calculate the kinetic energy in joules.

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

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

Broglie wavelength (λ) of the electron is calculated using the formula:

λ=h

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

the electron. Since p=√2mKE for an electron,

λ=h

√2mKE

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

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

λ=6.626 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−17 J

Step 5: Calculate the de Broglie wavelength.

λ=6.626 ×10−34 J·s

p2×9.11 ×10−31 kg ×1.6×10−17 J≈1.22 ×10−10 m

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

potential difference of 100 V is approximately 1.22 ×10−10 m.

Question 27

Question

A particle of mass mand charge qis moving with a velocity v. Determine its

de Broglie wavelength in terms of m,q, and v.

Solution

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

of a particle is given by:

λ=h

where λis the de Broglie wavelength, his the Planck constant, and pis the

momentum of the particle.

Step 2: Express momentum in terms of mass and velocity: The momentum

pof the particle is given by:

p=m·v

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

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

length equation: Substitute p=m·vinto the de Broglie wavelength equation:

λ=h

m·v

Step 4: Finalize the expression for the de Broglie wavelength: Therefore,

the de Broglie wavelength of the particle with mass m, charge qand velocity v

is:

λ=h

m·v

Question 28

Question

An electron with kinetic energy of 50 eV is incident on a crystal lattice. Cal-

culate the de Broglie wavelength of the electron and discuss the wave-particle

duality of electrons in this scenario.