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
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
