CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Wave-particle duality and de Broglie
wavelength
Question Bank - Set 4
Liberty University
Question 1
Question
A beam of electrons with momentum 3.0 x 10−24 kg m/s passes through a slit
of width 0.1 mm. Determine the de Broglie wavelength associated with these
electrons.
Solution
Step 1: Calculate the velocity of the electrons using the momentum formula
p=mv, where pis momentum, mis mass, and vis velocity.
Given: p= 3.0×10−24 kg m/s, m = 9.11 ×10−31 kg (mass of electron)
3.0×10−24 = (9.11 ×10−31)v
v=3.0×10−24
9.11 ×10−31 = 328946m/s
Step 2: Calculate the de Broglie wavelength using the formula λ=h
p, where
λis the de Broglie wavelength, his Planck’s constant ( 6.626 ×10−34 J s), and
pis momentum.
λ=6.626 ×10−34
3.0×10−24 = 2.21 ×10−10 m
Therefore, the de Broglie wavelength associated with these electrons is 2.21×
10−10 meters.
Question 2
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron in meters.
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. Step 2:
Calculate the velocity of the electron using the formula K=1
2mv2, where m
is the mass of the electron. Step 3: Calculate the de Broglie wavelength of the
electron using the formula λ=h
mv , where his the Planck constant.
Step 1: Calculate the kinetic energy of the electron.
K=eV = (1.6×10−19C)(100V)=1.6×10−17J
Step 2: Calculate the velocity of the electron.
K=1
2mv2
v=r2K
m=r2×1.6×10−17
9.11 ×10−31
v≈7.56 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron.
λ=h
mv =6.63 ×10−34Js
9.11 ×10−31kg ×7.56 ×106m/s
λ≈8.82 ×10−11 m
Therefore, the de Broglie wavelength of the electron is approximately 8.82 ×
10−11 meters.
Question 3
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after being accelerated. (Take the
charge of an electron as −e=−1.6×10−19 C and the mass of an electron as
9.11 ×10−31 kg)
2
Solution
Step 1: Firstly, calculate the kinetic energy of the electron using the formula:
KE =qV
where KE is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Step 2: Substitute the values into the formula to find the kinetic energy:
KE = (−1.6×10−19 C)(100 V)
Step 3: Solve for the kinetic energy:
KE =−1.6×10−17 J
Step 4: Next, use the kinetic energy to find the velocity of the electron using
the formula:
KE =1
2mv2
where mis the mass of the electron and vis its velocity.
Step 5: Rearrange the formula to solve for the velocity:
v=r2KE
m
Step 6: Substitute the values into the formula to find the velocity:
v=s2(−1.6×10−17 J)
9.11 ×10−31 kg
Step 7: Calculate the velocity of the electron.
Step 8: Finally, calculate the de Broglie wavelength of the electron using the
formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J
s), mis the mass of the electron, and vis its velocity.
Step 9: Substitute the values into the formula to find the de Broglie wave-
length:
λ=6.626 ×10−34 J s
(9.11 ×10−31 kg)(v)
Step 10: Substitute the calculated velocity into the formula and solve for
the de Broglie wavelength.
3
Question 4
Question
Consider an electron with kinetic energy E= 200 eV moving in a vacuum.
Determine the de Broglie wavelength associated with this electron.
Solution
To find the de Broglie wavelength of the electron, we can use the de Broglie
wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),
and pis the momentum.
Since the electron is moving, we can find its momentum using the formula:
E=p2
2m
where Eis the kinetic energy of the electron, pis the momentum, and mis the
mass of the electron (9.11 ×10−31 kg).
Step 1: Find the momentum of the electron.
p=√2mE
p=p2×9.11 ×10−31 kg ×1.6×10−19 J/eV ×200 eV
p≈2.606 ×10−24 kg m/s
Step 2: Calculate the de Broglie wavelength.
λ=6.626 ×10−34 m2kg/s
2.606 ×10−24 kg m/s
λ≈2.54 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 2.54 ×10−10 m.
Question 5
Question
An electron is accelerated through a potential difference of 1 keV. Determine
the de Broglie wavelength of the electron in meters.
4
Solution
Step 1: We start by finding the kinetic energy of the electron using the formula
KE =eV , where eis the elementary charge and Vis the potential difference.
Step 2: Substituting e= 1.6×10−19 C and V= 1000 V into the formula, we
get
KE = 1.6×10−19 C×1000 V = 1.6×10−16 J
Step 3: Next, we relate the kinetic energy of the electron to its de Broglie
wavelength. The de Broglie wavelength of a particle is given by the formula
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34
Js), and pis the momentum of the particle. The momentum of the electron can
be calculated using the formula
p=√2mE
where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy
of the electron. Step 4: Substituting the values of mand Einto the formula for
momentum, we get
p=p2×9.11 ×10−31 kg ×1.6×10−16 J
p≈5.6×10−25 kg m/s
Step 5: Finally, we substitute the values of hand pinto the formula for de
Broglie wavelength to find
λ=6.626 ×10−34 J s
5.6×10−25 kg m/s
λ≈1.18 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.18 ×
10−10 meters.
Question 6
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
5
Solution
Step 1: Find the kinetic energy of the electron using the formula KE =qV ,
where qis the charge of an electron and Vis the potential difference.
Given V= 100 V, q= 1.6×10−19 coulombs (charge of an electron)
Therefore, KE = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Use the kinetic energy to find the momentum of the electron. The
kinetic energy is equal to the energy associated with the particle’s motion:
KE =1
2mv2=p2
2mwhere p= momentum, m= mass of electron = 9.11×10−31 kg
Therefore, p=√2mKE =p2(9.11 ×10−31 kg)(1.6×10−17 J) = 3.82×10−24 kg m/s
Step 3: Use the momentum to find the de Broglie wavelength of the electron.
The de Broglie wavelength is given by λ=h
pwhere his the Planck constant.
Given h= 6.626 ×10−34 J s
Therefore, λ=6.626 ×10−34 J s
3.82 ×10−24 kg m/s = 1.74 ×10−10 m
Hence, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is 1.74 ×10−10 m.
Question 7
Question
A particle of mass mis moving with a velocity v. Determine the de Broglie
wavelength associated with this particle.
Solution
Step 1: The de Broglie wavelength is given by the formula:
λ=h
p
where his the Planck constant and pis the momentum of the particle.
Step 2: The momentum of the particle can be calculated using the formula:
p=mv
Step 3: Substituting the expression for momentum into the formula for de
Broglie wavelength, we get:
λ=h
mv
Step 4: Therefore, the de Broglie wavelength associated with a particle of
mass mmoving with velocity vis h
mv .
6
Question 8
Question
An electron with a kinetic energy of 200 eV is moving in vacuum. What is the
de Broglie wavelength of the electron?
Solution
Step 1: Convert the given kinetic energy into joules. Step 2: Calculate the speed
of the electron using the kinetic energy. Step 3: Use the de Broglie wavelength
formula to find the wavelength of the electron. Step 4: Express the wavelength
in standard units.
Step 1: Convert the given kinetic energy into joules. Given kinetic energy
of the electron = 200 eV
Since 1 eV = 1.6×10−19 J, the kinetic energy in joules is: 200 eV ×1.6×10−19
J/eV = 3.2×10−17 J.
Step 2: Calculate the speed of the electron using the kinetic energy. The
kinetic energy of the electron can be expressed as: K.E. =1
2mv2Where: -
K.E. is the kinetic energy, - mis the mass of the electron, - vis the speed of
the electron.
Solving for v:v=q2×K.E.
m
The mass of an electron, m= 9.11 ×10−31 kg. Substitute the values:
v=q2×3.2×10−17
9.11×10−31 v7.65 ×106m/s.
Step 3: Use the de Broglie wavelength formula to find the wavelength of
the electron. The de Broglie wavelength is given by: λ=h
pWhere: - λis the
wavelength, - his the Planck’s constant, - pis the momentum of the electron.
The momentum pof an electron is given by: p=mv
Substituting the values: p= 9.11 ×10−31kg ×7.65 ×106m/s p = 6.96 ×
10−24kg ·m/s
Then, calculate the de Broglie wavelength: λ=6.63×10−34 J·s
6.96×10−24 kg·m/s λ9.52 ×
10−11 m.
Step 4: Express the wavelength in standard units. Converting the wave-
length from meters to nanometers to get the final answer: λ9.52 ×10−11 m
= 95.2 nm.
Therefore, the de Broglie wavelength of the electron with a kinetic energy
of 200 eV is approximately 95.2 nm.
Question 9
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron.
7
Solution
Step 1: Calculate 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.
Step 1: Calculate the kinetic energy of the electron.
KE =eV = (1.6×10−19C)(100V)=1.6×10−17J
Step 2: Determine the momentum of the electron.
p=√2mKE =p2(9.11 ×10−31kg)(1.6×10−17 J)
p≈3.46 ×10−24kg ·m/s
Step 3: Calculate the de Broglie wavelength of the electron.
λ=h
p=6.63 ×10−34J·s
3.46 ×10−24kg ·m/s
λ≈1.91 ×10−10m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 100 V is approximately 1.91 ×10−10 m.
Question 10
Question
An electron with a speed of 2.5×106m/s is moving in a circular orbit. Calculate
the de Broglie wavelength associated with the electron.
Solution
Step 1: Calculate the momentum of the electron using the formula p=mv,
where mis the mass of the electron and vis its speed.
Given: v= 2.5×106m/s
Mass of an electron, m= 9.11 ×10−31 kg
p= (9.11 ×10−31 kg) ×(2.5×106m/s)
p= 2.2775 ×10−24 kg m/s
Step 2: Calculate the de Broglie wavelength using the formula λ=h
p, where
his the Planck constant.
Planck constant, h= 6.626 ×10−34 J s
8
λ=6.626 ×10−34 J s
2.2775 ×10−24 kg m/s
λ≈2.91 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 2.91 ×10−10 m.
Question 11
Question
An electron is accelerated through a potential difference of 150 V. Calculate the
de Broglie wavelength associated with the electron’s motion.
Solution
Step 1: Firstly, we need to find the kinetic energy of the electron using the
formula
Kinetic energy (KE) = Charge of electron ×Potential difference
KE = 1.6×10−19 ×150
KE = 2.4×10−17 J
Step 2: Next, we use the formula for kinetic energy in terms of momentum
to find the momentum of the electron:
Kinetic energy (KE) = 1
2×mass of electron ×(velocity of electron)2=p2
2m
Here, pis the momentum of the electron and mis the mass of the electron.
p=√2m×KE
p=p2×9.11 ×10−31 ×2.4×10−17
p≈5.43 ×10−25 kg m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
de Broglie wavelength = h
p
where his the Planck’s constant.
de Broglie wavelength = 6.626 ×10−34
5.43 ×10−25
de Broglie wavelength ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron’s motion
is approximately 1.22 ×10−10 m.
9
Question 12
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 potential difference.
Step 2: Use the kinetic energy to calculate the velocity of the electron. Step 3:
Apply the de Broglie wavelength formula to find the wavelength of the electron.
Step 1: Calculate the kinetic energy of the electron. The kinetic energy K
of the electron can be calculated using the formula:
K=eV
where eis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (100 V).
K= (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Calculate the velocity of the electron. The kinetic energy Kis
related to the velocity vof the electron by the formula:
K=1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg). Solving for v:
v=r2K
m=s2(1.6×10−17 J)
9.11 ×10−31 kg
v≈2.19 ×106m/s
Step 3: Apply the de Broglie wavelength formula. The de Broglie wave-
length λof the electron is given by:
λ=h
p=h
mv
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
(9.11 ×10−31 kg)(2.19 ×106m/s)
λ≈1.6×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 100 V is approximately 1.6×10−10 m.
10
Question 13
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Kinetic energy (KE) = e·V
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(100 V).
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron.
Kinetic energy (KE) = 1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg) and vis the velocity of
the electron.
1.6×10−17 =1
2(9.11 ×10−31)v2
v2=2×1.6×10−17
9.11 ×10−31
v=r3.2×10−17
9.11 ×10−31
v= 5.92 ×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the electron.
λ=h
mv
where his the Planck constant (6.626×10−34 J·s), mis the mass of the electron,
and vis the velocity of the electron calculated above.
λ=6.626 ×10−34
(9.11 ×10−31)(5.92 ×106)
λ=6.626 ×10−34
5.41 ×10−24
λ≈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.
11
Question 14
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron after acceleration. Given:
Charge of an electron, e=−1.6×10−19 C
Planck’s constant, h= 6.63 ×10−34 J-s
Mass of an electron, m= 9.11 ×10−31 kg
Speed of light, c= 3 ×108m/s
Solution
Step 1: Calculate the kinetic energy of the electron after acceleration using the
formula: KE =qV , where qis the charge and Vis the potential difference.
KE =e×V=−1.6×10−19 C×100 V = −1.6×10−17 J
Step 2: Calculate the velocity of the electron using the formula for kinetic
energy: KE =1
2mv2.
v=r2KE
m=s2× −1.6×10−17 J
9.11 ×10−31 kg ≈1.74 ×106m/s
Step 3: Now, calculate the de Broglie wavelength of the electron using the
formula: λ=h
p, where p=mv is the momentum of the electron.
λ=h
mv =6.63 ×10−34 J s
9.11 ×10−31 kg ×1.74 ×106m/s ≈7.15 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 100 V is approximately 7.15 ×10−10 m.
Question 15
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
12
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
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.
Step 1: Calculate the kinetic energy of the electron using the potential
difference. The potential energy gained by the electron is equal to its kinetic
energy. 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). Therefore, the kinetic energy (Ek) of the electron is:
Ek=qV = (−1.6×10−19 C)(100 V) = −1.6×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy of the electron can be expressed in terms of its velocity (v) using
the formula:
Ek=1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg). Substitute the known
values:
−1.6×10−17 =1
2(9.11 ×10−31)(v2)
Solving for vgives:
v=r−2(−1.6×10−17)
9.11 ×10−31 ≈5.93 ×106m/s
Step 3: Apply the de Broglie wavelength formula to find the 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 particle. The momentum of the electron is given by p=mv. Substitute the
values of mand vto find the momentum of the electron:
p= (9.11 ×10−31)(5.93 ×106)≈5.40 ×10−24 kg ·m/s
Now, calculate the de Broglie wavelength:
λ=6.63 ×10−34
5.40 ×10−24 ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.23 ×10−10 meters.
13
Question 16
Question
An electron with a kinetic energy of 50 eV is accelerated through a potential
difference of 100 V. Calculate the de Broglie wavelength associated with this
electron.
Solution
Step 1: Determine the velocity of the electron using its kinetic energy. Step 2:
Calculate the de Broglie wavelength using the velocity of the electron.
Step 1: The kinetic energy of the electron is given by the formula:
KE =1
2mv2
where KE is the kinetic energy, mis the mass of the electron, and vis the
velocity of the electron.
Given that the kinetic energy KE = 50 eV and the mass of the electron
m= 9.11 ×10−31 kg, we can rearrange the formula to solve for v:
v=r2KE
m=r2×50 ×1.6×10−19
9.11 ×10−31
Calculating the velocity gives:
v≈6.55 ×106m/s
Step 2: The de Broglie wavelength is given by the formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34
Js), mis the mass of the electron, and vis the velocity of the electron.
Substitute the values into the formula to find the de Broglie wavelength:
λ=6.63 ×10−34
9.11 ×10−31 ×6.55 ×106
Calculating the de Broglie wavelength gives:
λ≈1.46 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron is approx-
imately 1.46 ×10−10 meters.
14
Question 17
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron. Given that the charge
of an electron e= 1.6×10−19 C and Planck’s constant h= 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Step 2: Use the classical kinetic energy equation to find the velocity of the
electron. Step 3: Use the de Broglie wavelength formula to determine the de
Broglie wavelength.
Step 1: Calculate the kinetic energy of the electron using the potential
difference. The potential energy gained by the electron equals its kinetic energy.
Potential energy gained = Kinetic energy
e·V=1
2mv2
v=r2·e·V
m
Step 2: Use the classical kinetic energy equation to find the velocity of the
electron. The classical kinetic energy of the electron is given by:
Kinetic energy = 1
2mv2
Substitute the values and solve for the velocity:
v=r2·e·V
m
Step 3: Use the de Broglie wavelength formula to determine the de Broglie
wavelength. The de Broglie wavelength of a particle is given by:
λ=h
p=h
mv
Substitute the values of h,mand v:
λ=h
√2·e·V·m
λ=6.63 ×10−34
√2·1.6×10−19 ·100 ·9.11 ×10−31
λ≈6.63 ×10−34
√2·1.6×10−17 ·9.11 ×10−27
15
λ≈6.63 ×10−34
√2×1.4576 ×10−9×10−27
λ≈6.63 ×10−34
√2×1.4576 ×10−36
λ≈6.63 ×10−34
√2.9152 ×10−36
λ≈6.63 ×10−34
5.4×10−18
λ≈1.225 ×10−15
5.4×10−18
λ≈2.2685 ×10−3meters
Therefore, the de Broglie wavelength associated with the electron is approxi-
mately 2.27 ×10−3meters.
Question 18
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Begin by calculating the kinetic energy of the electron using the formula
K.E. =qV , where qis the charge of the electron (1.6×10−19 C) and Vis the
potential difference (100 V). Step 2: Substitute the values into the formula:
K.E. = (1.6×10−19 C)(100 V)
Step 3: Calculate the kinetic energy:
K.E. = 1.6×10−17 J
Step 4: Next, use the de Broglie wavelength formula λ=h
p, where his the
Planck constant (6.626 ×10−34 J·s) and pis the momentum of the electron.
Step 5: The momentum of the electron can be calculated using the formula
p=√2mK.E., where mis the mass of the electron (9.11 ×10−31 kg). Step 6:
Substitute the kinetic energy into the momentum formula:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
Step 7: Calculate the momentum of the electron:
p≈3.41 ×10−24 kg m/s
16
Step 8: Now, substitute the values of Planck’s constant and momentum into
the de Broglie wavelength formula:
λ=6.626 ×10−34 J·s
3.41 ×10−24 kg m/s
Step 9: Calculate the de Broglie wavelength:
λ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a po-
tential difference of 100 V is approximately 1.94 ×10−10 m.
Question 19
Question
An electron is accelerated through a potential difference of 50 V. Determine the
de Broglie wavelength associated with the electron.
Solution
The de Broglie wavelength associated with a particle can be calculated 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 particle.
The momentum of a particle can be calculated using the formula:
p=√2mE
where - mis the mass of the electron (9.11 ×10−31 kg), and - Eis the energy
of the particle.
Given that the potential difference is 50 V, we can calculate the kinetic
energy of the electron as:
E=eV
where - eis the elementary charge (1.6×10−19 C), - Vis the potential difference
(50 V).
Step 1: Calculating the kinetic energy
E=e×V= 1.6×10−19 C×50 V = 8 ×10−18 J
Step 2: Calculating the momentum
p=√2mE =p2×9.11 ×10−31 kg ×8×10−18 J
17
p≈p1.4568 ×10−17 ≈1.2075 ×10−8kg m/s
Step 3: Calculating the de Broglie wavelength
λ=h
p=6.626 ×10−34 m2kg/s
1.2075 ×10−8kg m/s
λ≈5.483 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 5.483 ×10−10 m.
Question 20
Question
An electron is accelerated through a potential difference of 200 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Use the energy of the electron to find its kinetic energy The kinetic
energy of the electron can be calculated using the equation:
Ke=eV
where Ke= kinetic energy, e= electron charge (1.6×10−19 C), V= potential
difference (200 V). Substitute the values and calculate:
Ke= (1.6×10−19 C)(200 V) = 3.2×10−17J
Step 2: Use the kinetic energy to find the de Broglie wavelength of the
electron The de Broglie wavelength of the electron can be calculated using the
equation:
λ=h
p
where λ= de Broglie wavelength, h= Planck’s constant (6.626 ×10−34 J·s),
p= momentum. The momentum of the electron can be calculated using the
equation:
p=p2mKe
where m= mass of the electron (9.11 ×10−31 kg). Substitute the values and
calculate the momentum:
p=p2(9.11 ×10−31 kg)(3.2×10−17J) = 9.05 ×10−23 kg ·m/s
Step 3: Calculate the de Broglie wavelength Now, substitute the momentum
value into the de Broglie wavelength equation and solve for λ:
λ=h
p=6.626 ×10−34 J·s
9.05 ×10−23 kg ·m/s = 7.32 ×10−11 m
18
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 200 V is 7.32 ×10−11 m.
Question 21
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 find the kinetic energy of the electron using the formula 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
Step 2: Next, we can find the velocity of the electron using the formula for
kinetic energy: KE =1
2mv2, where mis the mass of the electron.
1.6×10−17 =1
2(9.11 ×10−31)(v2)
v=r2(1.6×10−17)
9.11 ×10−31 ≈6.33 ×106m/s
Step 3: Now, we can calculate the de Broglie wavelength using the formula
λ=h
p, where λis the de Broglie wavelength, his the Planck constant, and pis
the momentum of the electron.
p=mv = (9.11 ×10−31 kg)(6.33 ×106m/s) ≈5.77 ×10−24 kg m/s
λ=6.626 ×10−34 J s
5.77 ×10−24 kg m/s ≈1.15 ×10−10 m = 115 pm
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 115 pm.
Question 22
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron after passing through
the potential difference.
19
Solution
Step 1: Find the kinetic energy of the electron using the potential difference.
Given that the potential difference is 100 V, we can use the equation K.E. =qV ,
where qis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (100 V).
K.E. = (1.6×10−19 C)×(100 V) = 1.6×10−17 J
Step 2: Use the kinetic energy to find the momentum of the electron. The
de Broglie wavelength is related to the momentum of a particle by the equa-
tion 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. First, calcu-
late the momentum using p=√2mK.E., where mis the mass of an electron
(9.11 ×10−31 kg).
p=p2×(9.11 ×10−31 kg)×(1.6×10−17 J) = p2.91 ×10−14 kg ·m/s
Step 3: Calculate the de Broglie wavelength. Now that we have the momen-
tum of the electron, we can find the de Broglie wavelength using the equation
λ=h
p.
λ=6.626 ×10−34 J·s
√2.91 ×10−14kg ·m/s ≈6.626 ×10−34
5.39 ×10−7m
λ≈1.23 ×10−10m
Therefore, the de Broglie wavelength associated with the electron after pass-
ing through the potential difference of 100 V is approximately 1.23 ×10−10 m.
Question 23
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula E=1
2mv2,
where mis the mass of an electron and vis its velocity.
Given that the potential difference V= 120 V, we know that the kinetic
energy KE gained by the electron is equal to the potential energy it loses, i.e.,
KE =qV , where qis the charge of the electron.
The charge of an electron is e= 1.6×10−19 C.
Therefore, KE =eV = (1.6×10−19 C)(120 V).
Step 2: Next, calculate the speed of the electron using the equation KE =
1
2mv2.
20
The mass of an electron m= 9.11 ×10−31 kg.
Substitute the known values into the equation: 1.6×10−19 ×120 = 1
2×
9.11 ×10−31 ×v2.
Step 3: Solve for vto find the speed of the electron.
Step 4: Now, calculate the de Broglie wavelength of the electron using the
formula λ=h
p, where his the Planck constant and pis the momentum of the
electron.
The momentum pof the electron can be calculated as p=mv.
Substitute the values of mand vto find p.
Step 5: Finally, calculate the de Broglie wavelength using the found values
of Planck constant hand momentum pin the formula λ=h
p.
Question 24
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Use the energy of the electron The kinetic energy of the electron can be
calculated using the potential difference and the charge of an electron (e).
K.E. =e·V
K.E. = (1.6×10−19 C) ·(120 V)
K.E. = 1.92 ×10−17 J
Step 2: Use the kinetic energy to find the momentum The kinetic energy
can be related to the momentum of the electron using the de Broglie wavelength
formula:
K.E. =p2
2m
where pis the momentum and mis the mass of the electron.
p=√2·m·K.E.
p=p2×(9.11 ×10−31 kg) ×(1.92 ×10−17 J)
p≈9.12 ×10−25 kg m/s
Step 3: Use the momentum to find the de Broglie wavelength The de Broglie
wavelength can be calculated using the momentum and the Planck constant (h).
λ=h
p
21
λ=6.63 ×10−34 J s
9.12 ×10−25 kg m/s
λ≈7.27 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 120 V is approximately 7.27 ×10−11 m.
Question 25
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Recall the de Broglie wavelength formula, which relates the wavelength
of a particle to its momentum:
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 of the electron, we can use the formula for
the kinetic energy of the electron:
K.E. =eV
where K.E. is the kinetic energy, eis the charge of an electron (1.6×10−19 C),
and Vis the potential difference.
Step 3: The kinetic energy of the electron can be expressed in terms of its
momentum pas:
K.E. =p2
2me
where meis the mass of the electron (9.11 ×10−31 kg).
Step 4: Setting the two expressions for kinetic energy equal to each other,
we have:
eV =p2
2me
Step 5: Rearranging the equation for momentum p, we get:
p=p2me·eV
Step 6: Substitute the values of me,e, and Vinto the equation to find the
momentum p.
Step 7: Now, substitute the calculated momentum pinto the de Broglie
wavelength formula to find the wavelength λassociated with the electron.
Step 8: Calculate the de Broglie wavelength λ.
22
Question 26
Question
Suppose an electron is accelerated through a potential difference of 100 V. Cal-
culate the de Broglie wavelength of the electron after acceleration. The mass of
an electron is 9.11 ×10−31 kg and the charge of an electron is 1.6×10−19 C.
Solution
Step 1: Use the energy conservation principle to find the kinetic energy of the
electron after acceleration. The potential energy that is converted into kinetic
energy can be calculated as follows:
P E =qV
P E = (1.6×10−19 C)(100V)
P E = 1.6×10−17 J
Step 2: The kinetic energy of the electron can be calculated using the for-
mula:
KE =P E =1
2mv2
where mis the mass of the electron, and vis the velocity of the electron. Since
the electron is accelerated from rest, the initial kinetic energy is 0, and the total
energy after acceleration is equal to the kinetic energy. Therefore:
KE = 1.6×10−17J
1
2mv2= 1.6×10−17J
v=s2×1.6×10−17J
9.11 ×10−31kg
Step 3: Once the velocity of the electron is found, the de Broglie wavelength
can be calculated using the formula:
λ=h
mv
where his the Planck constant. Substituting the values of h,m, and v:
λ=6.63 ×10−34J·s
(9.11 ×10−31kg)(q2×1.6×10−17 J
9.11×10−31 kg )
23
Question 27
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 potential difference.
Step 2: Use the de Broglie wavelength formula to find the wavelength.
Step 1: Calculate the kinetic energy of the electron. The potential energy
gained by the electron is equal to the kinetic energy gained. The kinetic energy
gained by the electron is given by the equation:
K.E. =qV
where qis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (200 V). Plugging in the values, we get:
K.E. = (1.6×10−19 C)(200 V)
K.E. = 3.2×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength is
given by the formula:
λ=h
p
where: - λis the de Broglie wavelength - his the Planck’s constant (6.626×10−34
J s) - pis the momentum of the electron, which can be calculated as p=
√2mK.E., where mis the mass of the electron (9.1×10−31 kg).
Plugging in the values for K.E. and m, we get:
p=p2(9.1×10−31 kg)(3.2×10−17 J)
p≈5.48 ×10−24 kg m/s
Therefore, the de Broglie wavelength is:
λ=6.626 ×10−34 J s
5.48 ×10−24 kg m/s
λ≈1.21 ×10−10 m
So, the de Broglie wavelength associated with the electron is approximately
1.21 ×10−10 m.
24
Question 28
Question
An electron with mass 9.11 ×10−31 kg is accelerated through a potential differ-
ence of 500 V. Calculate the de Broglie wavelength of the electron after accel-
eration.
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:
Calculate the velocity of the electron using the formula KE =1
2mv2, where m
is the mass of the electron and vis the velocity. Step 3: Calculate the de Broglie
wavelength using the formula λ=h
p, where his the Planck constant and pis
the momentum of the electron. Remember that p=mv for a non-relativistic
case.
Step 1: Given that q=−1.6×10−19 C and V= 500 V, we can calculate
the kinetic energy using the formula:
KE =qV = (−1.6×10−19 C)(500 V) = −8×10−17 J
Step 2: Now, let’s calculate the velocity of the electron using the kinetic
energy formula:
KE =1
2mv2
v=r2KE
m=s2(−8×10−17 J)
9.11 ×10−31 kg ≈5.75 ×106m/s
Step 3: Lastly, we can calculate the de Broglie wavelength using the for-
mula:
λ=h
p=h
mv
Plugging in the values h= 6.626 ×10−34 J s and m= 9.11 ×10−31 kg, we get:
λ=6.626 ×10−34 J s
(9.11 ×10−31 kg)(5.75 ×106m/s) ≈2.73 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 500 V is approximately 2.73 ×10−10 m.
Question 29
Question
An electron with a kinetic energy of 50 eV is incident on a crystal surface.
Assuming the electron behaves as a wave, calculate the de Broglie wavelength
of the electron.
25
Given: Planck’s constant, h= 6.626 ×10−34 J·s, charge of an electron,
e= 1.602 ×10−19 C, and the mass of an electron, m= 9.11 ×10−31 kg.
Solution
Step 1: Calculate the velocity of the electron using its kinetic energy.
Kinetic energy (KE) = 1
2mv2
v=r2·KE
m
v=r2×50 ×1.602 ×10−19
9.11 ×10−31
v≈1.93 ×106m/s
Step 2: Calculate the de Broglie wavelength of the electron using its velocity.
de Broglie wavelength = h
mv
λ=6.626 ×10−34
9.11 ×10−31 ×1.93 ×106
λ≈3.43 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 3.43 ×
10−10 m.
Question 30
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength of the electron following the acceleration. Recall that the
energy of an electron is given by E=qV , where qis the charge of the electron
and Vis the potential difference.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula E=qV .
The charge of an electron is q=−1.6×10−19 C. The potential difference is
V= 120 V. Substitute the values into the formula to find the energy:
E= (−1.6×10−19 C)(120 V)
E=−1.92 ×10−17 J
26
Step 2: Find the kinetic energy of the electron. The total energy of the
electron includes both its kinetic energy and its rest energy. The rest energy of
an electron is given by E0= 9.11 ×10−31 kg. The total energy is E=KE +E0,
where KE is the kinetic energy. Solving for KE, we get:
KE =E−E0
KE =−1.92 ×10−17 J−9.11 ×10−31 J
KE ≈ −1.92 ×10−17 J
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength is given by the formula λ=h
p, where his the Planck constant
6.63 ×10−34 Js, and pis the momentum of the electron. The momentum of
the electron can be calculated using p=√2mKE, where mis the mass of the
electron 9.11 ×10−31 kg. Substitute the values into the momentum formula:
p=p2(9.11 ×10−31 kg)(−1.92 ×10−17 J)
p=p−3.50 ×10−47 kg m/s
p=√3.50 ×10−24 kg m/s
The de Broglie wavelength is therefore:
λ=6.63 ×10−34 Js
√3.50 ×10−24 kg m/s
λ=6.63
√3.50 ×10−10 m
λ≈2.22 ×10−10 m
Question 31
Question
An electron is accelerated through a potential difference of 500 V. What is
the de Broglie wavelength associated with this electron? (Mass of electron
= 9.11 ×10−31 kg, charge of electron = −1.6×10−19 C, Planck’s constant
= 6.63 ×10−34 J s.)
Solution
Step 1: Calculate the kinetic energy of the electron by using the formula K.E. =
qV , where qis the charge and Vis the potential difference.
Step 1: K.E. = (1.6×10−19 C)(500 V)
Step 2: Calculate the de Broglie wavelength using the formula λ=h
p, where
pis the momentum of the electron (given by p=√2mK.E.).
Step 2: λ=6.63 ×10−34 J s
p2×9.11 ×10−31 kg ×(1.6×10−19 C×500 V)
27
Question 32
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 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 associated
with the electron.
Step 1: The kinetic energy of the electron can be calculated using the
formula K=eV , where eis the elementary charge and Vis the potential
difference. Given that e= 1.6×10−19 C and V= 200 V, the kinetic energy is:
K= (1.6×10−19 C)(200 V) = 3.2×10−17 J
Step 2: The momentum of the electron can be found using the formula
p=√2mK, where mis the mass of the electron. The mass of an electron,
m= 9.11 ×10−31 kg. Substituting the values, we get:
p=p2(9.11 ×10−31 kg)(3.2×10−17 J) ≈5.84 ×10−24 kg m/s
Step 3: The de Broglie wavelength is given by the formula λ=h
p, where his
the Planck constant. The Planck constant, h= 6.626 ×10−34 J s. Substituting
the values, we get:
λ=6.626 ×10−34 J s
5.84 ×10−24 kg m/s ≈1.13 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 200 V is approximately 1.13 ×10−10 m.
Question 33
Question
An electron is accelerated through a potential difference of 500 V. Determine
the de Broglie wavelength associated with this electron.
Solution
To determine the de Broglie wavelength of the electron accelerated through a
potential difference, we can use the formula relating energy, charge, potential
difference, and wavelength:
28
λ=h
√2meeV
where: - λ= de Broglie wavelength - h= Planck’s constant = 6.626 ×10−34
m2kg/s-me= mass of the electron = 9.11×10−31 kg - e= elementary charge
= 1.602 ×10−19 C - V= potential difference = 500 V
Step 1: Calculate the de Broglie wavelength using the formula.
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ≈6.626 ×10−34
√1.451 ×10−24 ≈6.626 ×10−34
1.205 ×10−12
λ≈6.626 ×10−22
1.205 ≈5.498 ×10−22 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a 500 V potential difference is approximately 5.498 ×10−22 m.
Question 34
Question
An electron is accelerated by a potential difference of 500 V. Calculate the de
Broglie wavelength of the electron.
Solution
Step 1: Given the potential difference V= 500 V, we can calculate the kinetic
energy Kof the electron using the relation K=eV , where eis the charge of an
electron (1.6×10−19 C).
K=eV = (1.6×10−19 C)(500 V) = 8 ×10−17 J
Step 2: The kinetic energy Kof the electron is related to its momentum p
by the equation K=p2
2m, where mis the mass of the electron (9.11 ×10−31 kg).
p=√2mK =p2(9.11 ×10−31 kg)(8 ×10−17 J) ≈9.50 ×10−24 kg m/s
29
Step 3: Using the de Broglie wavelength formula λ=h
p, where his the Planck
constant (6.63 ×10−34 m2kg/s), we can calculate the de Broglie wavelength of
the electron.
λ=6.63 ×10−34 m2kg/s
9.50 ×10−24 kg m/s ≈6.99 ×10−12 m
Therefore, the de Broglie wavelength of the electron accelerated by a poten-
tial difference of 500 V is approximately 6.99 ×10−12 m.
Question 35
Question
An electron is accelerated from rest through a potential difference of 100 V.
1. Calculate the final velocity of the electron in m/s.
2. Determine the de Broglie wavelength associated with this electron in nm.
Solution
1. Given that the potential difference is 100 V, we can find the final velocity
of the electron using the energy conservation principle. The change in
potential energy is equal to the gain in kinetic energy:
qV =1
2mv2
where qis the charge of the electron, Vis the potential difference, mis
the mass of the electron, and vis the final velocity.
Substituting the given values, q= 1.6×10−19 C, V= 100 V, and m=
9.11 ×10−31 kg, we can solve for v:
(1.6×10−19 C) ×(100 V) = 1
2(9.11 ×10−31 kg)v2
v=s2×(1.6×10−19 C) ×(100 V)
9.11 ×10−31 kg
v≈6.32 ×106m/s
Therefore, the final velocity of the electron is 6.32 ×106m/s.
2. The de Broglie wavelength of the electron can be calculated using the
following formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his Planck’s constant (6.626×10−34
J·s), mis the mass of the electron, vis the final velocity of the electron.
30
Question 2
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron in meters.
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. Step 2:
Calculate the velocity of the electron using the formula K=1
2mv2, where m
is the mass of the electron. Step 3: Calculate the de Broglie wavelength of the
electron using the formula λ=h
mv , where his the Planck constant.
Step 1: Calculate the kinetic energy of the electron.
K=eV = (1.6×10−19C)(100V)=1.6×10−17J
Step 2: Calculate the velocity of the electron.
K=1
2mv2
v=r2K
m=r2×1.6×10−17
9.11 ×10−31
v≈7.56 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron.
λ=h
mv =6.63 ×10−34Js
9.11 ×10−31kg ×7.56 ×106m/s
λ≈8.82 ×10−11 m
Therefore, the de Broglie wavelength of the electron is approximately 8.82 ×
10−11 meters.
Question 3
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after being accelerated. (Take the
charge of an electron as −e=−1.6×10−19 C and the mass of an electron as
9.11 ×10−31 kg)
2
Solution
Step 1: Firstly, calculate the kinetic energy of the electron using the formula:
KE =qV
where KE is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Step 2: Substitute the values into the formula to find the kinetic energy:
KE = (−1.6×10−19 C)(100 V)
Step 3: Solve for the kinetic energy:
KE =−1.6×10−17 J
Step 4: Next, use the kinetic energy to find the velocity of the electron using
the formula:
KE =1
2mv2
where mis the mass of the electron and vis its velocity.
Step 5: Rearrange the formula to solve for the velocity:
v=r2KE
m
Step 6: Substitute the values into the formula to find the velocity:
v=s2(−1.6×10−17 J)
9.11 ×10−31 kg
Step 7: Calculate the velocity of the electron.
Step 8: Finally, calculate the de Broglie wavelength of the electron using the
formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J
s), mis the mass of the electron, and vis its velocity.
Step 9: Substitute the values into the formula to find the de Broglie wave-
length:
λ=6.626 ×10−34 J s
(9.11 ×10−31 kg)(v)
Step 10: Substitute the calculated velocity into the formula and solve for
the de Broglie wavelength.
3
Question 4
Question
Consider an electron with kinetic energy E= 200 eV moving in a vacuum.
Determine the de Broglie wavelength associated with this electron.
Solution
To find the de Broglie wavelength of the electron, we can use the de Broglie
wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 m2kg/s),
and pis the momentum.
Since the electron is moving, we can find its momentum using the formula:
E=p2
2m
where Eis the kinetic energy of the electron, pis the momentum, and mis the
mass of the electron (9.11 ×10−31 kg).
Step 1: Find the momentum of the electron.
p=√2mE
p=p2×9.11 ×10−31 kg ×1.6×10−19 J/eV ×200 eV
p≈2.606 ×10−24 kg m/s
Step 2: Calculate the de Broglie wavelength.
λ=6.626 ×10−34 m2kg/s
2.606 ×10−24 kg m/s
λ≈2.54 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 2.54 ×10−10 m.
Question 5
Question
An electron is accelerated through a potential difference of 1 keV. Determine
the de Broglie wavelength of the electron in meters.
4
Solution
Step 1: We start by finding the kinetic energy of the electron using the formula
KE =eV , where eis the elementary charge and Vis the potential difference.
Step 2: Substituting e= 1.6×10−19 C and V= 1000 V into the formula, we
get
KE = 1.6×10−19 C×1000 V = 1.6×10−16 J
Step 3: Next, we relate the kinetic energy of the electron to its de Broglie
wavelength. The de Broglie wavelength of a particle is given by the formula
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34
Js), and pis the momentum of the particle. The momentum of the electron can
be calculated using the formula
p=√2mE
where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy
of the electron. Step 4: Substituting the values of mand Einto the formula for
momentum, we get
p=p2×9.11 ×10−31 kg ×1.6×10−16 J
p≈5.6×10−25 kg m/s
Step 5: Finally, we substitute the values of hand pinto the formula for de
Broglie wavelength to find
λ=6.626 ×10−34 J s
5.6×10−25 kg m/s
λ≈1.18 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.18 ×
10−10 meters.
Question 6
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
5
Solution
Step 1: Find the kinetic energy of the electron using the formula KE =qV ,
where qis the charge of an electron and Vis the potential difference.
Given V= 100 V, q= 1.6×10−19 coulombs (charge of an electron)
Therefore, KE = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Use the kinetic energy to find the momentum of the electron. The
kinetic energy is equal to the energy associated with the particle’s motion:
KE =1
2mv2=p2
2mwhere p= momentum, m= mass of electron = 9.11×10−31 kg
Therefore, p=√2mKE =p2(9.11 ×10−31 kg)(1.6×10−17 J) = 3.82×10−24 kg m/s
Step 3: Use the momentum to find the de Broglie wavelength of the electron.
The de Broglie wavelength is given by λ=h
pwhere his the Planck constant.
Given h= 6.626 ×10−34 J s
Therefore, λ=6.626 ×10−34 J s
3.82 ×10−24 kg m/s = 1.74 ×10−10 m
Hence, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is 1.74 ×10−10 m.
Question 7
Question
A particle of mass mis moving with a velocity v. Determine the de Broglie
wavelength associated with this particle.
Solution
Step 1: The de Broglie wavelength is given by the formula:
λ=h
p
where his the Planck constant and pis the momentum of the particle.
Step 2: The momentum of the particle can be calculated using the formula:
p=mv
Step 3: Substituting the expression for momentum into the formula for de
Broglie wavelength, we get:
λ=h
mv
Step 4: Therefore, the de Broglie wavelength associated with a particle of
mass mmoving with velocity vis h
mv .
6
Question 8
Question
An electron with a kinetic energy of 200 eV is moving in vacuum. What is the
de Broglie wavelength of the electron?
Solution
Step 1: Convert the given kinetic energy into joules. Step 2: Calculate the speed
of the electron using the kinetic energy. Step 3: Use the de Broglie wavelength
formula to find the wavelength of the electron. Step 4: Express the wavelength
in standard units.
Step 1: Convert the given kinetic energy into joules. Given kinetic energy
of the electron = 200 eV
Since 1 eV = 1.6×10−19 J, the kinetic energy in joules is: 200 eV ×1.6×10−19
J/eV = 3.2×10−17 J.
Step 2: Calculate the speed of the electron using the kinetic energy. The
kinetic energy of the electron can be expressed as: K.E. =1
2mv2Where: -
K.E. is the kinetic energy, - mis the mass of the electron, - vis the speed of
the electron.
Solving for v:v=q2×K.E.
m
The mass of an electron, m= 9.11 ×10−31 kg. Substitute the values:
v=q2×3.2×10−17
9.11×10−31 v7.65 ×106m/s.
Step 3: Use the de Broglie wavelength formula to find the wavelength of
the electron. The de Broglie wavelength is given by: λ=h
pWhere: - λis the
wavelength, - his the Planck’s constant, - pis the momentum of the electron.
The momentum pof an electron is given by: p=mv
Substituting the values: p= 9.11 ×10−31kg ×7.65 ×106m/s p = 6.96 ×
10−24kg ·m/s
Then, calculate the de Broglie wavelength: λ=6.63×10−34 J·s
6.96×10−24 kg·m/s λ9.52 ×
10−11 m.
Step 4: Express the wavelength in standard units. Converting the wave-
length from meters to nanometers to get the final answer: λ9.52 ×10−11 m
= 95.2 nm.
Therefore, the de Broglie wavelength of the electron with a kinetic energy
of 200 eV is approximately 95.2 nm.
Question 9
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron.
7
Solution
Step 1: Calculate 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.
Step 1: Calculate the kinetic energy of the electron.
KE =eV = (1.6×10−19C)(100V)=1.6×10−17J
Step 2: Determine the momentum of the electron.
p=√2mKE =p2(9.11 ×10−31kg)(1.6×10−17 J)
p≈3.46 ×10−24kg ·m/s
Step 3: Calculate the de Broglie wavelength of the electron.
λ=h
p=6.63 ×10−34J·s
3.46 ×10−24kg ·m/s
λ≈1.91 ×10−10m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 100 V is approximately 1.91 ×10−10 m.
Question 10
Question
An electron with a speed of 2.5×106m/s is moving in a circular orbit. Calculate
the de Broglie wavelength associated with the electron.
Solution
Step 1: Calculate the momentum of the electron using the formula p=mv,
where mis the mass of the electron and vis its speed.
Given: v= 2.5×106m/s
Mass of an electron, m= 9.11 ×10−31 kg
p= (9.11 ×10−31 kg) ×(2.5×106m/s)
p= 2.2775 ×10−24 kg m/s
Step 2: Calculate the de Broglie wavelength using the formula λ=h
p, where
his the Planck constant.
Planck constant, h= 6.626 ×10−34 J s
8
λ=6.626 ×10−34 J s
2.2775 ×10−24 kg m/s
λ≈2.91 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 2.91 ×10−10 m.
Question 11
Question
An electron is accelerated through a potential difference of 150 V. Calculate the
de Broglie wavelength associated with the electron’s motion.
Solution
Step 1: Firstly, we need to find the kinetic energy of the electron using the
formula
Kinetic energy (KE) = Charge of electron ×Potential difference
KE = 1.6×10−19 ×150
KE = 2.4×10−17 J
Step 2: Next, we use the formula for kinetic energy in terms of momentum
to find the momentum of the electron:
Kinetic energy (KE) = 1
2×mass of electron ×(velocity of electron)2=p2
2m
Here, pis the momentum of the electron and mis the mass of the electron.
p=√2m×KE
p=p2×9.11 ×10−31 ×2.4×10−17
p≈5.43 ×10−25 kg m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
de Broglie wavelength = h
p
where his the Planck’s constant.
de Broglie wavelength = 6.626 ×10−34
5.43 ×10−25
de Broglie wavelength ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron’s motion
is approximately 1.22 ×10−10 m.
9
Question 12
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 potential difference.
Step 2: Use the kinetic energy to calculate the velocity of the electron. Step 3:
Apply the de Broglie wavelength formula to find the wavelength of the electron.
Step 1: Calculate the kinetic energy of the electron. The kinetic energy K
of the electron can be calculated using the formula:
K=eV
where eis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (100 V).
K= (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Calculate the velocity of the electron. The kinetic energy Kis
related to the velocity vof the electron by the formula:
K=1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg). Solving for v:
v=r2K
m=s2(1.6×10−17 J)
9.11 ×10−31 kg
v≈2.19 ×106m/s
Step 3: Apply the de Broglie wavelength formula. The de Broglie wave-
length λof the electron is given by:
λ=h
p=h
mv
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
(9.11 ×10−31 kg)(2.19 ×106m/s)
λ≈1.6×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 100 V is approximately 1.6×10−10 m.
10
Question 13
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Kinetic energy (KE) = e·V
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(100 V).
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron.
Kinetic energy (KE) = 1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg) and vis the velocity of
the electron.
1.6×10−17 =1
2(9.11 ×10−31)v2
v2=2×1.6×10−17
9.11 ×10−31
v=r3.2×10−17
9.11 ×10−31
v= 5.92 ×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the electron.
λ=h
mv
where his the Planck constant (6.626×10−34 J·s), mis the mass of the electron,
and vis the velocity of the electron calculated above.
λ=6.626 ×10−34
(9.11 ×10−31)(5.92 ×106)
λ=6.626 ×10−34
5.41 ×10−24
λ≈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.
11
Question 14
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron after acceleration. Given:
Charge of an electron, e=−1.6×10−19 C
Planck’s constant, h= 6.63 ×10−34 J-s
Mass of an electron, m= 9.11 ×10−31 kg
Speed of light, c= 3 ×108m/s
Solution
Step 1: Calculate the kinetic energy of the electron after acceleration using the
formula: KE =qV , where qis the charge and Vis the potential difference.
KE =e×V=−1.6×10−19 C×100 V = −1.6×10−17 J
Step 2: Calculate the velocity of the electron using the formula for kinetic
energy: KE =1
2mv2.
v=r2KE
m=s2× −1.6×10−17 J
9.11 ×10−31 kg ≈1.74 ×106m/s
Step 3: Now, calculate the de Broglie wavelength of the electron using the
formula: λ=h
p, where p=mv is the momentum of the electron.
λ=h
mv =6.63 ×10−34 J s
9.11 ×10−31 kg ×1.74 ×106m/s ≈7.15 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 100 V is approximately 7.15 ×10−10 m.
Question 15
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
12
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
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.
Step 1: Calculate the kinetic energy of the electron using the potential
difference. The potential energy gained by the electron is equal to its kinetic
energy. 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). Therefore, the kinetic energy (Ek) of the electron is:
Ek=qV = (−1.6×10−19 C)(100 V) = −1.6×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy of the electron can be expressed in terms of its velocity (v) using
the formula:
Ek=1
2mv2
where mis the mass of the electron (9.11 ×10−31 kg). Substitute the known
values:
−1.6×10−17 =1
2(9.11 ×10−31)(v2)
Solving for vgives:
v=r−2(−1.6×10−17)
9.11 ×10−31 ≈5.93 ×106m/s
Step 3: Apply the de Broglie wavelength formula to find the 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 particle. The momentum of the electron is given by p=mv. Substitute the
values of mand vto find the momentum of the electron:
p= (9.11 ×10−31)(5.93 ×106)≈5.40 ×10−24 kg ·m/s
Now, calculate the de Broglie wavelength:
λ=6.63 ×10−34
5.40 ×10−24 ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.23 ×10−10 meters.
13
Question 16
Question
An electron with a kinetic energy of 50 eV is accelerated through a potential
difference of 100 V. Calculate the de Broglie wavelength associated with this
electron.
Solution
Step 1: Determine the velocity of the electron using its kinetic energy. Step 2:
Calculate the de Broglie wavelength using the velocity of the electron.
Step 1: The kinetic energy of the electron is given by the formula:
KE =1
2mv2
where KE is the kinetic energy, mis the mass of the electron, and vis the
velocity of the electron.
Given that the kinetic energy KE = 50 eV and the mass of the electron
m= 9.11 ×10−31 kg, we can rearrange the formula to solve for v:
v=r2KE
m=r2×50 ×1.6×10−19
9.11 ×10−31
Calculating the velocity gives:
v≈6.55 ×106m/s
Step 2: The de Broglie wavelength is given by the formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34
Js), mis the mass of the electron, and vis the velocity of the electron.
Substitute the values into the formula to find the de Broglie wavelength:
λ=6.63 ×10−34
9.11 ×10−31 ×6.55 ×106
Calculating the de Broglie wavelength gives:
λ≈1.46 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron is approx-
imately 1.46 ×10−10 meters.
14
Question 17
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron. Given that the charge
of an electron e= 1.6×10−19 C and Planck’s constant h= 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Step 2: Use the classical kinetic energy equation to find the velocity of the
electron. Step 3: Use the de Broglie wavelength formula to determine the de
Broglie wavelength.
Step 1: Calculate the kinetic energy of the electron using the potential
difference. The potential energy gained by the electron equals its kinetic energy.
Potential energy gained = Kinetic energy
e·V=1
2mv2
v=r2·e·V
m
Step 2: Use the classical kinetic energy equation to find the velocity of the
electron. The classical kinetic energy of the electron is given by:
Kinetic energy = 1
2mv2
Substitute the values and solve for the velocity:
v=r2·e·V
m
Step 3: Use the de Broglie wavelength formula to determine the de Broglie
wavelength. The de Broglie wavelength of a particle is given by:
λ=h
p=h
mv
Substitute the values of h,mand v:
λ=h
√2·e·V·m
λ=6.63 ×10−34
√2·1.6×10−19 ·100 ·9.11 ×10−31
λ≈6.63 ×10−34
√2·1.6×10−17 ·9.11 ×10−27
15
λ≈6.63 ×10−34
√2×1.4576 ×10−9×10−27
λ≈6.63 ×10−34
√2×1.4576 ×10−36
λ≈6.63 ×10−34
√2.9152 ×10−36
λ≈6.63 ×10−34
5.4×10−18
λ≈1.225 ×10−15
5.4×10−18
λ≈2.2685 ×10−3meters
Therefore, the de Broglie wavelength associated with the electron is approxi-
mately 2.27 ×10−3meters.
Question 18
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Begin by calculating the kinetic energy of the electron using the formula
K.E. =qV , where qis the charge of the electron (1.6×10−19 C) and Vis the
potential difference (100 V). Step 2: Substitute the values into the formula:
K.E. = (1.6×10−19 C)(100 V)
Step 3: Calculate the kinetic energy:
K.E. = 1.6×10−17 J
Step 4: Next, use the de Broglie wavelength formula λ=h
p, where his the
Planck constant (6.626 ×10−34 J·s) and pis the momentum of the electron.
Step 5: The momentum of the electron can be calculated using the formula
p=√2mK.E., where mis the mass of the electron (9.11 ×10−31 kg). Step 6:
Substitute the kinetic energy into the momentum formula:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
Step 7: Calculate the momentum of the electron:
p≈3.41 ×10−24 kg m/s
16
Step 8: Now, substitute the values of Planck’s constant and momentum into
the de Broglie wavelength formula:
λ=6.626 ×10−34 J·s
3.41 ×10−24 kg m/s
Step 9: Calculate the de Broglie wavelength:
λ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a po-
tential difference of 100 V is approximately 1.94 ×10−10 m.
Question 19
Question
An electron is accelerated through a potential difference of 50 V. Determine the
de Broglie wavelength associated with the electron.
Solution
The de Broglie wavelength associated with a particle can be calculated 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 particle.
The momentum of a particle can be calculated using the formula:
p=√2mE
where - mis the mass of the electron (9.11 ×10−31 kg), and - Eis the energy
of the particle.
Given that the potential difference is 50 V, we can calculate the kinetic
energy of the electron as:
E=eV
where - eis the elementary charge (1.6×10−19 C), - Vis the potential difference
(50 V).
Step 1: Calculating the kinetic energy
E=e×V= 1.6×10−19 C×50 V = 8 ×10−18 J
Step 2: Calculating the momentum
p=√2mE =p2×9.11 ×10−31 kg ×8×10−18 J
17
p≈p1.4568 ×10−17 ≈1.2075 ×10−8kg m/s
Step 3: Calculating the de Broglie wavelength
λ=h
p=6.626 ×10−34 m2kg/s
1.2075 ×10−8kg m/s
λ≈5.483 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 5.483 ×10−10 m.
Question 20
Question
An electron is accelerated through a potential difference of 200 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Use the energy of the electron to find its kinetic energy The kinetic
energy of the electron can be calculated using the equation:
Ke=eV
where Ke= kinetic energy, e= electron charge (1.6×10−19 C), V= potential
difference (200 V). Substitute the values and calculate:
Ke= (1.6×10−19 C)(200 V) = 3.2×10−17J
Step 2: Use the kinetic energy to find the de Broglie wavelength of the
electron The de Broglie wavelength of the electron can be calculated using the
equation:
λ=h
p
where λ= de Broglie wavelength, h= Planck’s constant (6.626 ×10−34 J·s),
p= momentum. The momentum of the electron can be calculated using the
equation:
p=p2mKe
where m= mass of the electron (9.11 ×10−31 kg). Substitute the values and
calculate the momentum:
p=p2(9.11 ×10−31 kg)(3.2×10−17J) = 9.05 ×10−23 kg ·m/s
Step 3: Calculate the de Broglie wavelength Now, substitute the momentum
value into the de Broglie wavelength equation and solve for λ:
λ=h
p=6.626 ×10−34 J·s
9.05 ×10−23 kg ·m/s = 7.32 ×10−11 m
18
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 200 V is 7.32 ×10−11 m.
Question 21
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 find the kinetic energy of the electron using the formula 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
Step 2: Next, we can find the velocity of the electron using the formula for
kinetic energy: KE =1
2mv2, where mis the mass of the electron.
1.6×10−17 =1
2(9.11 ×10−31)(v2)
v=r2(1.6×10−17)
9.11 ×10−31 ≈6.33 ×106m/s
Step 3: Now, we can calculate the de Broglie wavelength using the formula
λ=h
p, where λis the de Broglie wavelength, his the Planck constant, and pis
the momentum of the electron.
p=mv = (9.11 ×10−31 kg)(6.33 ×106m/s) ≈5.77 ×10−24 kg m/s
λ=6.626 ×10−34 J s
5.77 ×10−24 kg m/s ≈1.15 ×10−10 m = 115 pm
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 115 pm.
Question 22
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron after passing through
the potential difference.
19
Solution
Step 1: Find the kinetic energy of the electron using the potential difference.
Given that the potential difference is 100 V, we can use the equation K.E. =qV ,
where qis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (100 V).
K.E. = (1.6×10−19 C)×(100 V) = 1.6×10−17 J
Step 2: Use the kinetic energy to find the momentum of the electron. The
de Broglie wavelength is related to the momentum of a particle by the equa-
tion 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. First, calcu-
late the momentum using p=√2mK.E., where mis the mass of an electron
(9.11 ×10−31 kg).
p=p2×(9.11 ×10−31 kg)×(1.6×10−17 J) = p2.91 ×10−14 kg ·m/s
Step 3: Calculate the de Broglie wavelength. Now that we have the momen-
tum of the electron, we can find the de Broglie wavelength using the equation
λ=h
p.
λ=6.626 ×10−34 J·s
√2.91 ×10−14kg ·m/s ≈6.626 ×10−34
5.39 ×10−7m
λ≈1.23 ×10−10m
Therefore, the de Broglie wavelength associated with the electron after pass-
ing through the potential difference of 100 V is approximately 1.23 ×10−10 m.
Question 23
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula E=1
2mv2,
where mis the mass of an electron and vis its velocity.
Given that the potential difference V= 120 V, we know that the kinetic
energy KE gained by the electron is equal to the potential energy it loses, i.e.,
KE =qV , where qis the charge of the electron.
The charge of an electron is e= 1.6×10−19 C.
Therefore, KE =eV = (1.6×10−19 C)(120 V).
Step 2: Next, calculate the speed of the electron using the equation KE =
1
2mv2.
20
The mass of an electron m= 9.11 ×10−31 kg.
Substitute the known values into the equation: 1.6×10−19 ×120 = 1
2×
9.11 ×10−31 ×v2.
Step 3: Solve for vto find the speed of the electron.
Step 4: Now, calculate the de Broglie wavelength of the electron using the
formula λ=h
p, where his the Planck constant and pis the momentum of the
electron.
The momentum pof the electron can be calculated as p=mv.
Substitute the values of mand vto find p.
Step 5: Finally, calculate the de Broglie wavelength using the found values
of Planck constant hand momentum pin the formula λ=h
p.
Question 24
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Use the energy of the electron The kinetic energy of the electron can be
calculated using the potential difference and the charge of an electron (e).
K.E. =e·V
K.E. = (1.6×10−19 C) ·(120 V)
K.E. = 1.92 ×10−17 J
Step 2: Use the kinetic energy to find the momentum The kinetic energy
can be related to the momentum of the electron using the de Broglie wavelength
formula:
K.E. =p2
2m
where pis the momentum and mis the mass of the electron.
p=√2·m·K.E.
p=p2×(9.11 ×10−31 kg) ×(1.92 ×10−17 J)
p≈9.12 ×10−25 kg m/s
Step 3: Use the momentum to find the de Broglie wavelength The de Broglie
wavelength can be calculated using the momentum and the Planck constant (h).
λ=h
p
21
λ=6.63 ×10−34 J s
9.12 ×10−25 kg m/s
λ≈7.27 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 120 V is approximately 7.27 ×10−11 m.
Question 25
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Recall the de Broglie wavelength formula, which relates the wavelength
of a particle to its momentum:
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 of the electron, we can use the formula for
the kinetic energy of the electron:
K.E. =eV
where K.E. is the kinetic energy, eis the charge of an electron (1.6×10−19 C),
and Vis the potential difference.
Step 3: The kinetic energy of the electron can be expressed in terms of its
momentum pas:
K.E. =p2
2me
where meis the mass of the electron (9.11 ×10−31 kg).
Step 4: Setting the two expressions for kinetic energy equal to each other,
we have:
eV =p2
2me
Step 5: Rearranging the equation for momentum p, we get:
p=p2me·eV
Step 6: Substitute the values of me,e, and Vinto the equation to find the
momentum p.
Step 7: Now, substitute the calculated momentum pinto the de Broglie
wavelength formula to find the wavelength λassociated with the electron.
Step 8: Calculate the de Broglie wavelength λ.
22
Question 26
Question
Suppose an electron is accelerated through a potential difference of 100 V. Cal-
culate the de Broglie wavelength of the electron after acceleration. The mass of
an electron is 9.11 ×10−31 kg and the charge of an electron is 1.6×10−19 C.
Solution
Step 1: Use the energy conservation principle to find the kinetic energy of the
electron after acceleration. The potential energy that is converted into kinetic
energy can be calculated as follows:
P E =qV
P E = (1.6×10−19 C)(100V)
P E = 1.6×10−17 J
Step 2: The kinetic energy of the electron can be calculated using the for-
mula:
KE =P E =1
2mv2
where mis the mass of the electron, and vis the velocity of the electron. Since
the electron is accelerated from rest, the initial kinetic energy is 0, and the total
energy after acceleration is equal to the kinetic energy. Therefore:
KE = 1.6×10−17J
1
2mv2= 1.6×10−17J
v=s2×1.6×10−17J
9.11 ×10−31kg
Step 3: Once the velocity of the electron is found, the de Broglie wavelength
can be calculated using the formula:
λ=h
mv
where his the Planck constant. Substituting the values of h,m, and v:
λ=6.63 ×10−34J·s
(9.11 ×10−31kg)(q2×1.6×10−17 J
9.11×10−31 kg )
23
Question 27
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 potential difference.
Step 2: Use the de Broglie wavelength formula to find the wavelength.
Step 1: Calculate the kinetic energy of the electron. The potential energy
gained by the electron is equal to the kinetic energy gained. The kinetic energy
gained by the electron is given by the equation:
K.E. =qV
where qis the charge of an electron (1.6×10−19 C) and Vis the potential
difference (200 V). Plugging in the values, we get:
K.E. = (1.6×10−19 C)(200 V)
K.E. = 3.2×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength is
given by the formula:
λ=h
p
where: - λis the de Broglie wavelength - his the Planck’s constant (6.626×10−34
J s) - pis the momentum of the electron, which can be calculated as p=
√2mK.E., where mis the mass of the electron (9.1×10−31 kg).
Plugging in the values for K.E. and m, we get:
p=p2(9.1×10−31 kg)(3.2×10−17 J)
p≈5.48 ×10−24 kg m/s
Therefore, the de Broglie wavelength is:
λ=6.626 ×10−34 J s
5.48 ×10−24 kg m/s
λ≈1.21 ×10−10 m
So, the de Broglie wavelength associated with the electron is approximately
1.21 ×10−10 m.
24
Question 28
Question
An electron with mass 9.11 ×10−31 kg is accelerated through a potential differ-
ence of 500 V. Calculate the de Broglie wavelength of the electron after accel-
eration.
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:
Calculate the velocity of the electron using the formula KE =1
2mv2, where m
is the mass of the electron and vis the velocity. Step 3: Calculate the de Broglie
wavelength using the formula λ=h
p, where his the Planck constant and pis
the momentum of the electron. Remember that p=mv for a non-relativistic
case.
Step 1: Given that q=−1.6×10−19 C and V= 500 V, we can calculate
the kinetic energy using the formula:
KE =qV = (−1.6×10−19 C)(500 V) = −8×10−17 J
Step 2: Now, let’s calculate the velocity of the electron using the kinetic
energy formula:
KE =1
2mv2
v=r2KE
m=s2(−8×10−17 J)
9.11 ×10−31 kg ≈5.75 ×106m/s
Step 3: Lastly, we can calculate the de Broglie wavelength using the for-
mula:
λ=h
p=h
mv
Plugging in the values h= 6.626 ×10−34 J s and m= 9.11 ×10−31 kg, we get:
λ=6.626 ×10−34 J s
(9.11 ×10−31 kg)(5.75 ×106m/s) ≈2.73 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 500 V is approximately 2.73 ×10−10 m.
Question 29
Question
An electron with a kinetic energy of 50 eV is incident on a crystal surface.
Assuming the electron behaves as a wave, calculate the de Broglie wavelength
of the electron.
25
Given: Planck’s constant, h= 6.626 ×10−34 J·s, charge of an electron,
e= 1.602 ×10−19 C, and the mass of an electron, m= 9.11 ×10−31 kg.
Solution
Step 1: Calculate the velocity of the electron using its kinetic energy.
Kinetic energy (KE) = 1
2mv2
v=r2·KE
m
v=r2×50 ×1.602 ×10−19
9.11 ×10−31
v≈1.93 ×106m/s
Step 2: Calculate the de Broglie wavelength of the electron using its velocity.
de Broglie wavelength = h
mv
λ=6.626 ×10−34
9.11 ×10−31 ×1.93 ×106
λ≈3.43 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 3.43 ×
10−10 m.
Question 30
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength of the electron following the acceleration. Recall that the
energy of an electron is given by E=qV , where qis the charge of the electron
and Vis the potential difference.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula E=qV .
The charge of an electron is q=−1.6×10−19 C. The potential difference is
V= 120 V. Substitute the values into the formula to find the energy:
E= (−1.6×10−19 C)(120 V)
E=−1.92 ×10−17 J
26
Step 2: Find the kinetic energy of the electron. The total energy of the
electron includes both its kinetic energy and its rest energy. The rest energy of
an electron is given by E0= 9.11 ×10−31 kg. The total energy is E=KE +E0,
where KE is the kinetic energy. Solving for KE, we get:
KE =E−E0
KE =−1.92 ×10−17 J−9.11 ×10−31 J
KE ≈ −1.92 ×10−17 J
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength is given by the formula λ=h
p, where his the Planck constant
6.63 ×10−34 Js, and pis the momentum of the electron. The momentum of
the electron can be calculated using p=√2mKE, where mis the mass of the
electron 9.11 ×10−31 kg. Substitute the values into the momentum formula:
p=p2(9.11 ×10−31 kg)(−1.92 ×10−17 J)
p=p−3.50 ×10−47 kg m/s
p=√3.50 ×10−24 kg m/s
The de Broglie wavelength is therefore:
λ=6.63 ×10−34 Js
√3.50 ×10−24 kg m/s
λ=6.63
√3.50 ×10−10 m
λ≈2.22 ×10−10 m
Question 31
Question
An electron is accelerated through a potential difference of 500 V. What is
the de Broglie wavelength associated with this electron? (Mass of electron
= 9.11 ×10−31 kg, charge of electron = −1.6×10−19 C, Planck’s constant
= 6.63 ×10−34 J s.)
Solution
Step 1: Calculate the kinetic energy of the electron by using the formula K.E. =
qV , where qis the charge and Vis the potential difference.
Step 1: K.E. = (1.6×10−19 C)(500 V)
Step 2: Calculate the de Broglie wavelength using the formula λ=h
p, where
pis the momentum of the electron (given by p=√2mK.E.).
Step 2: λ=6.63 ×10−34 J s
p2×9.11 ×10−31 kg ×(1.6×10−19 C×500 V)
27
Question 32
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 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 associated
with the electron.
Step 1: The kinetic energy of the electron can be calculated using the
formula K=eV , where eis the elementary charge and Vis the potential
difference. Given that e= 1.6×10−19 C and V= 200 V, the kinetic energy is:
K= (1.6×10−19 C)(200 V) = 3.2×10−17 J
Step 2: The momentum of the electron can be found using the formula
p=√2mK, where mis the mass of the electron. The mass of an electron,
m= 9.11 ×10−31 kg. Substituting the values, we get:
p=p2(9.11 ×10−31 kg)(3.2×10−17 J) ≈5.84 ×10−24 kg m/s
Step 3: The de Broglie wavelength is given by the formula λ=h
p, where his
the Planck constant. The Planck constant, h= 6.626 ×10−34 J s. Substituting
the values, we get:
λ=6.626 ×10−34 J s
5.84 ×10−24 kg m/s ≈1.13 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 200 V is approximately 1.13 ×10−10 m.
Question 33
Question
An electron is accelerated through a potential difference of 500 V. Determine
the de Broglie wavelength associated with this electron.
Solution
To determine the de Broglie wavelength of the electron accelerated through a
potential difference, we can use the formula relating energy, charge, potential
difference, and wavelength:
28
λ=h
√2meeV
where: - λ= de Broglie wavelength - h= Planck’s constant = 6.626 ×10−34
m2kg/s-me= mass of the electron = 9.11×10−31 kg - e= elementary charge
= 1.602 ×10−19 C - V= potential difference = 500 V
Step 1: Calculate the de Broglie wavelength using the formula.
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ=6.626 ×10−34
√2×9.11 ×10−31 ×1.602 ×10−19 ×500
λ≈6.626 ×10−34
√1.451 ×10−24 ≈6.626 ×10−34
1.205 ×10−12
λ≈6.626 ×10−22
1.205 ≈5.498 ×10−22 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a 500 V potential difference is approximately 5.498 ×10−22 m.
Question 34
Question
An electron is accelerated by a potential difference of 500 V. Calculate the de
Broglie wavelength of the electron.
Solution
Step 1: Given the potential difference V= 500 V, we can calculate the kinetic
energy Kof the electron using the relation K=eV , where eis the charge of an
electron (1.6×10−19 C).
K=eV = (1.6×10−19 C)(500 V) = 8 ×10−17 J
Step 2: The kinetic energy Kof the electron is related to its momentum p
by the equation K=p2
2m, where mis the mass of the electron (9.11 ×10−31 kg).
p=√2mK =p2(9.11 ×10−31 kg)(8 ×10−17 J) ≈9.50 ×10−24 kg m/s
29
Step 3: Using the de Broglie wavelength formula λ=h
p, where his the Planck
constant (6.63 ×10−34 m2kg/s), we can calculate the de Broglie wavelength of
the electron.
λ=6.63 ×10−34 m2kg/s
9.50 ×10−24 kg m/s ≈6.99 ×10−12 m
Therefore, the de Broglie wavelength of the electron accelerated by a poten-
tial difference of 500 V is approximately 6.99 ×10−12 m.
Question 35
Question
An electron is accelerated from rest through a potential difference of 100 V.
1. Calculate the final velocity of the electron in m/s.
2. Determine the de Broglie wavelength associated with this electron in nm.
Solution
1. Given that the potential difference is 100 V, we can find the final velocity
of the electron using the energy conservation principle. The change in
potential energy is equal to the gain in kinetic energy:
qV =1
2mv2
where qis the charge of the electron, Vis the potential difference, mis
the mass of the electron, and vis the final velocity.
Substituting the given values, q= 1.6×10−19 C, V= 100 V, and m=
9.11 ×10−31 kg, we can solve for v:
(1.6×10−19 C) ×(100 V) = 1
2(9.11 ×10−31 kg)v2
v=s2×(1.6×10−19 C) ×(100 V)
9.11 ×10−31 kg
v≈6.32 ×106m/s
Therefore, the final velocity of the electron is 6.32 ×106m/s.
2. The de Broglie wavelength of the electron can be calculated using the
following formula:
λ=h
p=h
mv
where λis the de Broglie wavelength, his Planck’s constant (6.626×10−34
J·s), mis the mass of the electron, vis the final velocity of the electron.
30
Substituting the known values into the formula, we get:
λ=6.626 ×10−34 J·s
(9.11 ×10−31 kg) ×(6.32 ×106m/s)
λ≈1.23 ×10−10 m
Converting the wavelength to nanometers, we have:
λ≈123 nm
Therefore, the de Broglie wavelength associated with the electron is ap-
proximately 123 nm.
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