CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Wave-particle duality and de Broglie
wavelength
Question Bank - Set 5
Liberty University
Question 1
Question
An electron with a mass of 9.11 ×10−31 kg is accelerated through a potential
difference of 120 V. Calculate the de Broglie wavelength of the electron after
acceleration.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =qV ,
where qis the charge of an electron and Vis the potential difference.
Step 1: KE = (1.6×10−19 C)(120 V) = 1.92 ×10−17 J
Step 2: Using the formula for kinetic energy KE =1
2mv2and the formula
for de Broglie wavelength λ=h
pwhere his the Planck constant and pis the
momentum, solve for v.
Step 2: 1.92 ×10−17 =1
2×(9.11 ×10−31)×v2
Step 2 cont.: v=r2×1.92 ×10−17
9.11 ×10−31 = 6.67 ×106m/s
Step 3: Calculate the momentum of the electron using p=mv.
Step 3: p= (9.11 ×10−31 kg)(6.67 ×106m/s) = 6.07 ×10−24 kg m/s
Step 4: Calculate the de Broglie wavelength of the electron using the formula
λ=h
p.
Step 4: λ=6.63 ×10−34 Js
6.07 ×10−24 kg m/s = 1.09 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
1.09 ×10−10 m.
Question 2
Question
An electron with a kinetic energy of 200 eV is accelerated through a potential
difference of 50 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the velocity of the electron using the kinetic energy.
Kinetic energy (KE) = 1
2mv2
Given that the mass of an electron m= 9.11 ×10−31 kg and the kinetic energy
KE = 200 eV, where 1 eV = 1.6×10−19 J, we have:
200 ×1.6×10−19 =1
2×9.11 ×10−31 ×v2
v=r2×200 ×1.6×10−19
9.11 ×10−31
Step 2: Substitute the value of vinto the de Broglie wavelength formula.
λ=h
p=h
mv
where h= 6.63 ×10−34 J s is the Planck’s constant.
Step 3: Calculate the de Broglie wavelength. Substitute m,v, and hinto
the formula and solve for λ.
λ=6.63 ×10−34
9.11 ×10−31 ×v
Question 3
Question
A particle is moving with a velocity of 1.5×106m/s. Calculate the de Broglie
wavelength associated with this particle.
2
Solution
Step 1: Recall the de Broglie wavelength formula:
λ=h
p
where: λ= de Broglie wavelength (m), h= Planck’s constant (6.626×10−34 m2kg/s),
p= momentum of the particle (kg m/s).
Step 2: First, we need to determine the momentum of the particle using the
formula:
p=mv
where: m= mass of the particle, v= velocity of the particle.
Step 3: Let’s assume the particle is an electron with mass 9.11 ×10−31 kg.
Calculate the momentum using the mass and velocity given:
p= (9.11 ×10−31 kg) ×(1.5×106m/s)
Step 4: Calculate the momentum of the electron:
p= 1.365 ×10−24 kg m/s
Step 5: Now, substitute the momentum into the de Broglie wavelength for-
mula:
λ=6.626 ×10−34 m2kg/s
1.365 ×10−24 kg m/s
Step 6: Calculate the de Broglie wavelength:
λ= 4.853 ×10−10 m
Therefore, the de Broglie wavelength associated with the particle moving at
1.5×106m/s is 4.853 ×10−10 m.
Question 4
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Given that the potential difference is 100 V, we can calculate the kinetic
energy of the electron using the equation K.E. =eV , where eis the charge of
an electron (1.6×10−19 C) and Vis the potential difference.
3
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
Step 2: Use the kinetic energy to find the de Broglie wavelength of the
electron.
The de Broglie wavelength of a particle is given by the equation λ=h
p,
where his the Planck constant (6.626 ×10−34 J·s) and pis the momentum of
the electron.
Since p=√2mK.E. for a non-relativistic particle, where mis the mass of
the electron (9.11 ×10−31 kg),
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J)
p≈3.027 ×10−24 kg ·m/s
Therefore, the de Broglie wavelength of the electron is:
λ=6.626 ×10−34 J·s
3.027 ×10−24 kg ·m/s
λ≈2.19 ×10−10 m
Question 5
Question
A particle with mass mand speed vexhibits wave-particle duality. Determine
its de Broglie wavelength λin terms of mand v.
Solution
To determine the de Broglie wavelength λof a particle with mass mand speed
v, we can 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 particle.
Step 1: First, we need to find the momentum pof the particle in terms of
its mass mand speed v. The momentum pis given by:
p=mv
4
Step 2: Now, substitute p=mv into the de Broglie wavelength formula:
λ=h
mv
So, the de Broglie wavelength λin terms of the mass mand speed vis
λ=h
mv .
Question 6
Question
A particle with mass mand velocity vis accelerated through a potential differ-
ence V. Determine the de Broglie wavelength associated with this particle.
Solution
Step 1: Determine the kinetic energy of the particle. The kinetic energy Kof
the particle can be calculated using the formula:
K=1
2mv2
Step 2: Relate the potential difference with the kinetic energy. The potential
energy Vcan be converted into the kinetic energy through the equation:
eV =K
where eis the elementary charge.
Step 3: Express the velocity in terms of potential difference. From steps
1 and 2, we can relate the velocity vto the potential difference Vusing the
equations:
K=1
2mv2
eV =K
Step 4: Calculate the de Broglie wavelength. The de Broglie wavelength λ
associated with the particle can be found using the equation:
λ=h
p
where his the Planck’s constant and pis the momentum of the particle.
Step 5: Express the momentum in terms of mass and velocity. The momen-
tum pof the particle can be related to its mass mand velocity vthrough the
equation:
p=mv
5
Step 6: Substitute the expression for momentum into the equation for de
Broglie wavelength. Substitute p=mv into the equation λ=h
pto get:
λ=h
mv
Step 7: Substitute the velocity expression in terms of potential difference into
the de Broglie wavelength equation. Substitute v=q2eV
minto the equation
λ=h
mv to get the de Broglie wavelength associated with the particle accelerated
through a potential difference V.
Question 7
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 (1.6 x 10−19 C) and Vis the potential
difference (100 V).
KE = (1.6×10−19 C) ×(100 V) = 1.6×10−17 J
Step 2: Next, we can find the momentum of the electron using the formula
p=√2me·KE, where meis the mass of the electron (9.11 x 10−31 kg).
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J) ≈3.49 ×10−24 kg m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
λ=h
p, where his the Planck constant (6.63 x 10−34 J s).
λ=6.63 ×10−34 J s
3.49 ×10−24 kg m/s ≈1.90 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.90 ×10−10 m.
Question 8
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
6
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =eV ,
where eis the charge of an electron (1.6×10−19 C) and Vis the potential
difference. Step 2: Calculate the velocity of the electron using the formula
KE =1
2mv2, where mis the mass of an electron (9.11 ×10−31 kg). Step 3:
Calculate the de Broglie wavelength using the formula λ=h
mv , where his the
Planck constant (6.63 ×10−34 J s).
Step 1: The kinetic energy KE of the electron is given by
KE =eV = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Using the kinetic energy formula 1
2mv2=KE, we have
1
2(9.11 ×10−31 kg)v2= 1.6×10−17 J
v2=2×1.6×10−17
9.11 ×10−31
v2= 3.5133 ×1013 m2/s2
v≈5.92 ×106m/s
Step 3: The de Broglie wavelength λis given by
λ=h
mv =6.63 ×10−34 J s
9.11 ×10−31 kg ×5.92 ×106m/s
λ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron acceler-
ated through a potential difference of 100 V is 1.22 ×10−10 m.
Question 9
Question
A beam of electrons with a kinetic energy of 500 eV strikes a crystal. If the
crystal diffracts the electrons at an angle of 30◦relative to the incident direction,
what is the de Broglie wavelength of the electrons?
Solution
Step 1: Calculate the speed of the electrons using their kinetic energy. Step 2:
Use the speed of the electrons to find their momentum. Step 3: Apply the de
Broglie wavelength formula to determine the wavelength of the electrons.
7
Step 1: Calculate the speed of the electrons. The kinetic energy of the
electrons is given by the formula:
KE =1
2mv2
Here, KE = 500 eV = 500 ×1.6×10−19 J(1eV=1.6×10−19 J), and we know
the mass of an electron m= 9.11 ×10−31 kg. Solving for v:
v=r2KE
m=r2×500 ×1.6×10−19
9.11 ×10−31
Step 2: Use the speed of the electrons to find their momentum. The mo-
mentum of an electron is given by:
p=mv
Substitute the values of mand vinto the above equation to find p.
Step 3: Apply the de Broglie wavelength formula to determine the wave-
length of the electrons. The de Broglie wavelength of the electrons is given
by:
λ=h
p
where h= 6.626 ×10−34 J s is the Planck constant. Substitute the calculated
value of pinto the de Broglie wavelength formula to find the wavelength of the
electrons.
Question 10
Question
An electron is accelerated through a potential difference of 100 V. Determine
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 formula:
K.E. =qV =eV
where qis the charge of the electron, Vis the potential difference, and eis the
elementary charge. Substituting the given values:
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
8
Step 2: Use the kinetic energy to find the de Broglie wavelength. The de
Broglie wavelength of a particle can be calculated using the formula:
λ=h
p
where his the Planck constant and pis the momentum of the electron. The
momentum of the electron can be calculated using the formula:
p=√2mK.E.
where mis the mass of the electron. Substituting the given values:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈3.42 ×10−24 kg m/s
Step 3: Substitute the momentum into the de Broglie wavelength formula.
λ=6.626 ×10−34 J s
3.42 ×10−24 kg m/s
λ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.94 ×
10−10 m.
Question 11
Question
An electron is accelerated through a potential difference of 1000 V. Calculate
the de Broglie wavelength of the electron after acceleration. The mass of an
electron is 9.11 ×10−31 kg and the elementary charge is 1.60 ×10−19 C.
Solution
Step 1: Calculate the kinetic energy of the electron. Step 2: Use the de Broglie
wavelength formula to find the wavelength of the electron.
Step 1: Calculate the kinetic energy of the electron.
The potential energy gained by the electron as it is accelerated through the
potential difference is equal to its kinetic energy. The kinetic energy of the
electron can be calculated using the formula:
KE =q·V
Where: KE = kinetic energy, q= charge of the electron, and V= potential
difference.
Substitute the values into the formula:
9
KE = (1.60 ×10−19 C)·(1000 V)
KE = 1.60 ×10−16 J
Therefore, the kinetic energy of the electron is 1.60 ×10−16 J.
Step 2: Use the de Broglie wavelength formula to find the wavelength of
the electron.
The de Broglie wavelength of an electron is given by the formula:
λ=h
p
Where: λ= de Broglie wavelength, h= Planck’s constant (6.63 ×10−34 J
s), p= momentum.
The momentum of the electron can be calculated using the formula:
p=√2mKE
Where: m= mass of the electron, and KE = kinetic energy.
Substitute the values into the formula:
p=p2×(9.11 ×10−31 kg)×(1.60 ×10−16 J)
p≈4.23 ×10−23 kg ·m/s
Now, substitute the momentum into the de Broglie wavelength formula:
λ=6.63 ×10−34 Js
4.23 ×10−23 kg ·m/s
λ≈1.57 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.57 ×10−10 m.
Question 12
Question
An electron is accelerated by a potential difference of 200 V. Calculate the de
Broglie wavelength of the electron.
10
Solution
Step 1: We first need to find the kinetic energy of the electron using the for-
mula K.E. =qV , where qis the charge of the electron and Vis the potential
difference.
K.E. =qV
= (1.6×10−19 C)(200 V)
= 3.2×10−17 J
Step 2: Next, we will use the equation for de Broglie wavelength, λ=h
p,
where his the Planck constant (6.626 ×10−34 J·s), pis the momentum of the
electron, and λis the de Broglie wavelength. We can find the momentum of the
electron using the formula p=√2mK.E., where mis the mass of the electron.
p=√2mK.E.
=p2×9.11 ×10−31 kg ×3.2×10−17 J
= 8.53 ×10−13 kg ·m/s
Step 3: Now, we can compute the de Broglie wavelength of the electron:
λ=h
p
=6.626 ×10−34 J·s
8.53 ×10−13 kg ·m/s
= 7.77 ×10−11 m
Therefore, the de Broglie wavelength of the electron accelerated by a poten-
tial difference of 200 V is 7.77 ×10−11 meters.
Question 13
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with the electron.
(Note: The charge of an electron is −1.6×10−19 C and its mass is 9.11×10−31
kg.)
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =qV ,
where qis the charge and Vis the potential difference.
KE = (1.6×10−19 C)(120 V)
= 1.92 ×10−17 J
11
Step 2: Use the kinetic energy to calculate the speed of the electron using
the kinetic energy formula KE =1
2mv2, where mis the mass of the electron
and vis the speed.
1.92 ×10−17 =1
2(9.11 ×10−31)v2
v=r2×1.92 ×10−17
9.11 ×10−31
v≈5.94 ×106m/s
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 (p=mv).
λ=6.626 ×10−34 J s
9.11 ×10−31 kg ×5.94 ×106m/s
λ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with the accelerated electron
is approximately 1.22 ×10−10 m.
Question 14
Question
An electron with a kinetic energy of 500 eV exhibits wave-like behavior. Deter-
mine the de Broglie wavelength of the electron.
Solution
Step 1: Convert the kinetic energy of the electron from electron volts to joules.
Step 2: Use the de Broglie wavelength formula to calculate the wavelength of
the electron.
Step 1: To convert the kinetic energy from electron volts (eV) to joules (J),
we use the conversion factor 1 eV = 1.6×10−19 J:
Kinetic energy = 500 eV ×1.6×10−19 J/eV = 8 ×10−17 J
Step 2: The de Broglie wavelength (λ) of a particle with momentum pis
given by the formula:
λ=h
p
where his the Planck constant (6.63 ×10−34 J·s) and pis the momentum of
the particle. The momentum of an electron can be calculated using its kinetic
energy:
p=√2mE
12
where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy
(in joules).
Substitute the values into the formula:
p=p2×9.11 ×10−31 kg ×8×10−17 J≈5.35 ×10−24 kg m/s
Now, substitute hand pinto the de Broglie wavelength formula:
λ=6.63 ×10−34 J·s
5.35 ×10−24 kg m/s ≈1.24 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.24 ×
10−10 m.
Question 15
Question
An electron with kinetic energy of 200 eV is accelerated through a potential
difference of 50 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: First, we need to convert the kinetic energy of the electron from electron
volts to joules. We know that 1 eV is equivalent to 1.6×10−19 J. So, the kinetic
energy in joules is given by:
KE = 200 ×1.6×10−19 = 3.2×10−17 J
Step 2: Next, we can find the velocity of the electron using the kinetic energy
formula: KE =1
2mv2, where mis the mass of the electron and vis its velocity.
The mass of an electron is approximately 9.11 ×10−31 kg. Thus, we have:
3.2×10−17 =1
2×9.11 ×10−31 ×v2
v=r2×3.2×10−17
9.11 ×10−31
v≈5.36 ×106m/s
Step 3: Now, we can calculate the de Broglie wavelength of the electron
using its momentum, p=mv, and the de Broglie wavelength formula: λ=h
p,
where his the Planck constant. The momentum of the electron is given by:
p= 9.11 ×10−31 ×5.36 ×106= 4.88 ×10−24 kg m/s
Thus, the de Broglie wavelength is:
λ=6.626 ×10−34
4.88 ×10−24
13
λ≈1.36 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.36 ×
10−10 meters.
Question 16
Question
An electron is accelerated through a potential difference of 100 V. What is
the de Broglie wavelength of the electron after acceleration? Given that the
elementary charge of an electron is e= 1.6×10−19 C, the mass of an electron
is me= 9.11 ×10−31 kg, and Planck’s constant is h= 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron. The kinetic energy of an
electron accelerated through a potential difference ∆Vis given by the equation:
K.E. =e·∆V
Substitute the values for the elementary charge of an electron and the potential
difference into the equation:
K.E. = (1.6×10−19 C) ·(100 V) = 1.6×10−17 J
Step 2: Calculate the de Broglie wavelength using the kinetic energy. The de
Broglie wavelength of a particle with kinetic energy K.E. and mass mis given
by the equation:
λ=h
√2mK.E.
Substitute the values for Planck’s constant, the mass of an electron, and the
kinetic energy of the electron into the equation:
λ=6.63 ×10−34 J·s
p2·(9.11 ×10−31 kg) ·(1.6×10−17 J)
λ=6.63 ×10−34 J·s
p2·(9.11 ×10−14)
λ=6.63 ×10−34 J·s
√1.82 ×10−13
λ=6.63 ×10−34 J·s
1.35 ×10−6
λ= 4.89 ×10−28 m
Therefore, the de Broglie wavelength of the electron after acceleration is
4.89 ×10−28 m.
14
Question 17
Question
A photon with energy 3.0×10−19 J is incident on a metal surface. Calculate
the de Broglie wavelength of an electron ejected from the metal surface if the
work function of the metal is 4.2×10−19 J.
Solution
Step 1: Calculate the kinetic energy of the ejected electron. Given that the
energy of the incident photon is 3.0×10−19 J, and the work function of the
metal is 4.2×10−19 J, the kinetic energy (K.E.) of the ejected electron can be
calculated as:
K.E. =Ephoton −Work Function
K.E. = 3.0×10−19 −4.2×10−19
K.E. =−1.2×10−19 J
Step 2: Calculate the speed of the ejected electron using the kinetic energy.
The kinetic energy of the electron can be related to its speed using the equation
K.E. =1
2mv2, where mis the mass of an electron.
1
2mv2=−1.2×10−19
v2=−2× −1.2×10−19
9.1×10−31
v≈4.71 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength of an electron can be calculated using the formula λ=h
mv , where h
is Planck’s constant.
λ=h
mv =6.63 ×10−34
9.1×10−31 ×4.71 ×106
λ≈1.51 ×10−10 m
Therefore, the de Broglie wavelength of the ejected electron is approximately
1.51 ×10−10 m.
Question 18
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron.
15
Solution
Step 1: Calculate the kinetic energy of the electron using the formula K.E. =
qV , where qis the charge of the electron and Vis the potential difference.
Step 1: K.E. =qV = (1.6×10−19 C)(500 V) = 8 ×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron using the
formula K.E. =1
2mv2, where mis the mass of the electron.
Step 2: 1
2mv2= 8×10−17 J⇒v=r2K.E.
m=s2(8 ×10−17 J)
9.11 ×10−31 kg ≈1.49×106m/s
Step 3: 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.
Step 3: p=mv = (9.11 ×10−31 kg)(1.49 ×106m/s) ≈1.362 ×10−24 kg m/s
λ=h
p=6.63 ×10−34 J s
1.362 ×10−24 kg m/s ≈4.87 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 500 V is approximately 4.87 ×10−10 m.
Question 19
Question
A particle of mass mand velocity vexhibits wave-like behavior according to
wave-particle duality. Calculate the de Broglie wavelength of the particle in
terms of its momentum.
Solution
To determine the de Broglie wavelength of a particle in terms of its momentum,
we can use the de Broglie wavelength formula:
λ=h
p
where: - λis the de Broglie wavelength, - his the Planck constant (6.62607015×
10−34m2kg/s), - pis the momentum of the particle.
Step 1: Determine the momentum of the particle using its mass
and velocity The momentum of a particle is given by the formula:
p=m·v
where: - mis the mass of the particle, - vis the velocity of the particle.
16
Step 2: Substitute the momentum into the de Broglie wavelength
formula Substitute the expression for momentum (p=m·v) into the de Broglie
wavelength formula:
λ=h
m·v
Step 3: Simplify the expression The de Broglie wavelength can be sim-
plified to:
λ=h
m·v
Therefore, the de Broglie wavelength of the particle in terms of its momen-
tum is λ=h
m·v.
Question 20
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron after acceleration. (The charge of an
electron is −1.6×10−19 C and the mass of an electron is 9.11 ×10−31 kg.)
Solution
Step 1: Calculate the kinetic energy of the electron using the potential differ-
ence. The potential energy gained by the electron is equal to the kinetic energy
gained. Given that the charge of an electron is −1.6×10−19 C and the poten-
tial difference is 100 V, we can use the formula for potential energy: P E =qV .
Substitute the values in:
P E =−1.6×10−19 ×100 = −1.6×10−17 J
Since the potential energy gained equals the kinetic energy gained, the kinetic
energy of the electron is 1.6×10−17 J.
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy of the electron is given by:
KE =1
2mv2
where mis the mass of the electron. Substitute the values in and solve for v:
1.6×10−17 =1
2×9.11 ×10−31 ×v2
v≈5.352 ×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the electron.
The de Broglie wavelength of a particle is given by:
λ=h
mv
17
where his the Planck constant (6.63 ×10−34 J s). Substitute the values in:
λ=6.63 ×10−34
9.11 ×10−31 ×5.352 ×106
λ≈1.224 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.224 ×10−10 m.
Question 21
Question
An electron is accelerated through a potential difference of 200 V. Calculate the
de Broglie wavelength associated with the electron after acceleration.
Solution
We can use the de Broglie wavelength formula to find the wavelength associated
with 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 particle.
To find the momentum of the electron, we first need to find its kinetic energy
using the given potential difference. The kinetic energy of the electron is given
by:
KE =eV
where eis the charge of an electron (1.6×10−19 C), and Vis the potential
difference.
Step 1: Calculate the kinetic energy of the electron.
KE =eV = (1.6×10−19 C)(200 V)
KE = 3.2×10−17 J
Step 2: Calculate the momentum of the electron. The kinetic energy can
also be expressed in terms of the momentum p:
KE =p2
2m
where mis the mass of the electron (9.11 ×10−31 kg).
3.2×10−17 =p2
2(9.11 ×10−31)
18
p2= 6.4×10−17 ×2×9.11 ×10−31
p=p1.16384 ×10−30
p≈1.078 ×10−15
Step 3: Calculate the de Broglie wavelength.
λ=h
p=6.626 ×10−34
1.078 ×10−15
λ≈6.14 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 6.14 ×10−10 m.
Question 22
Question
An electron with a kinetic energy of 150 eV is incident on a single slit with a
width of 0.1 mm. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: We will first convert the given kinetic energy of the electron from electron
volts (eV) to joules (J). Given: E= 150 eV
Recall that 1 eV is equivalent to 1.602 ×10−19 J. Therefore, we have:
E= 150 ×1.602 ×10−19 J
Step 2: Now, we can calculate the velocity of the electron using the kinetic
energy formula:
E=1
2mv2
where mis the mass of the electron and vis the velocity. Since the mass of
an electron is 9.11 ×10−31 kg, we can solve for v.
Step 3: Next, we will calculate the de Broglie wavelength using the 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 electron.
Step 4: To find the momentum of the electron, we use the formula:
p=mv
Step 5: Finally, we substitute the calculated values of pand hinto the de
Broglie wavelength formula to find the wavelength of the electron.
19
Question 23
Question
An electron is accelerated from rest through a potential difference of 120 V. Cal-
culate the de Broglie wavelength associated with the electron after acceleration.
(Hint: The de Broglie wavelength λof a particle is given by λ=h
p, where h
is the Planck constant (6.626 ×10−34 m2kg/s) and pis the momentum of the
particle. For an electron, p=mv, where m= 9.11 ×10−31 kg is the mass of
the electron and vis the final velocity of the electron after acceleration. Use
the fact that the kinetic energy gained by the electron equals the work done on
it by the potential difference, i.e., eV =1
2mv2.)
Solution
Step 1: The kinetic energy gained by the electron is given by eV , where eis
the charge of an electron (1.6×10−19 C) and Vis the potential difference (120
V). This energy is equal to the kinetic energy of the electron after acceleration,
which is 1
2mv2. Setting eV =1
2mv2:
eV =1
2mv2
1.6×10−19 ×120 = 1
2×9.11 ×10−31 ×v2
1.92 ×10−17 = 4.55 ×10−31 ×v2
v2=1.92 ×10−17
4.55 ×10−31
v2≈4.23 ×1013 m/s
Step 2: The momentum pof the electron is equal to mv:
p=mv
p= 9.11 ×10−31 ×4.23 ×1013
p≈3.86 ×10−17 kg m/s
Step 3: Now, we can calculate the de Broglie wavelength λusing the formula
λ=h
p:
λ=h
p
λ=6.626 ×10−34
3.86 ×10−17
λ≈1.72 ×10−16 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 1.72 ×10−16 meters.
20
Question 24
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: First, we need to calculate the kinetic energy of the electron using the
formula K.E. =e·V, where eis the elementary charge and Vis the potential
difference. Given that the potential difference is 500 V, and the elementary
charge is 1.6×10−19 C, we have:
K.E. = 1.6×10−19 C×500 V
K.E. = 8 ×10−17 J
Step 2: Next, we can calculate the velocity of the electron using the for-
mula K.E. =1
2mv2, where mis the mass of the electron (9.11 x 10−31kg).8×
10−17 J = 1
2×9.11 ×10−31 kg ×v2
v=s2×8×10−17 J
9.11 ×10−31 kg
v= 714936.78 m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
λ=h
p, where his the Planck’s constant (6.63 x 10−34J.s)andp is the momentum
of the electron (m·v).
λ=6.63 ×10−34 J.s
9.11 ×10−31 kg ×714936.78 m/s
λ=6.63 ×10−34 J.s
6.51 ×10−24 kg.m/s
λ≈1.02 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron is approx-
imately 1.02 ×10−10 m.
Question 25
Question
A proton is accelerated by a potential difference of 100 V. Calculate the de
Broglie wavelength of the proton using the formula λ=h
p, where λis the de
Broglie wavelength, his the Planck constant (6.63 ×10−34 m2kg/s), and pis
the momentum of the proton.
21
Solution
Step 1: Calculate the momentum of the proton using the formula p=√2mE,
where mis the mass of the proton and Eis the kinetic energy acquired by
the proton. Given that the charge of the proton, e= 1.6×10−19 C, and the
potential difference, V= 100 V, the kinetic energy can be calculated as:
E=eV = (1.6×10−19 C)(100 V)
Step 2: Substitute the kinetic energy into the formula p=√2mE to find
the momentum:
p=p2×(1.67 ×10−27 kg) ×E
Step 3: Calculate the de Broglie wavelength using the momentum obtained
in step 2:
λ=h
p
Question 26
Question
An electron is accelerated through a potential difference of 100 V. Calculate 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 his Planck’s constant (6.626 ×10−34 m2kg/s) and pis the momentum of
the electron.
Step 1: Find the kinetic energy of the electron The kinetic energy of
an electron accelerated through a potential difference ∆Vis given by:
K.E. =e·∆V
where eis the charge of an electron (1.6×10−19 C) and ∆Vis the potential
difference through which the electron is accelerated. Substituting the values,
we get:
K.E. = (1.6×10−19 C) ·(100 V) = 1.6×10−17 J
Step 2: Find the momentum of the electron The momentum of the
electron can be calculated using the formula:
p=√2mK.E.
22
where mis the mass of the electron (9.11 ×10−31 kg) and K.E. is the kinetic
energy of the electron. Substituting the values, we get:
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J) ≈3.41 ×10−24 kg m/s
Step 3: Calculate the de Broglie wavelength Now, we can calculate
the de Broglie wavelength using the formula:
λ=h
p
Substitute the values of hand pto find the de Broglie wavelength:
λ=6.626 ×10−34 m2kg/s
3.41 ×10−24 kg m/s ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.94 ×10−10 m.
Question 27
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron.
Solution
Step 1: We can find the kinetic energy of the electron using the formula K.E. =
qV , where qis the charge of the electron and Vis the potential difference. Given
V= 100 V and charge of an electron q= 1.6×10−19 C,
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
Step 2: The de Broglie wavelength of a particle is given by the formula
λ=h
p, where his the Planck constant and pis the momentum. The momentum
of the electron can be calculated using p=√2mK.E., where mis the mass
of the electron and K.E. is the kinetic energy. Given mass of electron m=
9.11 ×10−31 kg,
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈3.4×10−24 kg m/s
Step 3: Now, we can find the de Broglie wavelength:
λ=h
p=6.63 ×10−34 J s
3.4×10−24 kg m/s
23
λ≈1.95 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.95 ×10−10 m.
Question 28
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Let’s start by using the kinetic energy equation for an electron accelerated
through a potential difference:
K=eV
where Kis the kinetic energy, eis the elementary charge (1.6×10−19 C), and
Vis the potential difference.
Step 1: Calculate the kinetic energy of the electron.
K=eV
K= (1.6×10−19 C)(100 V)
K= 1.6×10−17 J
Step 2: Use the de Broglie wavelength formula to find the wavelength
associated with the electron:
λ=h
p=h
√2mK
where his the Planck constant (6.63 ×10−34 J·s), mis the mass of the electron
(9.11 ×10−31 kg), and Kis the kinetic energy.
Step 3: Substitute the values into the formula to find the de Broglie wave-
length.
λ=h
√2mK
λ=6.63 ×10−34 J·s
p2×9.11 ×10−31 kg ×1.6×10−17 J
λ≈6.63 ×10−34
5.39 ×10−24 m
λ≈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 m.
24
Question 29
Question
An electron with a kinetic energy of 200 eV is traveling at a speed of 2 ×106
m/s. Determine:
1. The de Broglie wavelength of the electron.
2. Whether the electron exhibits wave-like or particle-like behavior.
Solution
1. To find the de Broglie wavelength of the electron, we can use the de Broglie
wavelength formula:
λ=h
p
where
p=mv
and mis the mass of the electron, vis its speed, his the Planck constant, and
λis the de Broglie wavelength.
Given:
Kinetic energy, K= 200 eV = 200 ×1.6×10−19 J
Speed, v= 2 ×106m/s
Mass of electron, m= 9.11 ×10−31 kg
Planck constant, h= 6.63 ×10−34 J s
First, calculate the momentum p:
p=mv
= (9.11 ×10−31 kg)(2 ×106m/s)
= 1.822 ×10−24 kg m/s
Now, calculate the de Broglie wavelength λ:
λ=h
p
=6.63 ×10−34 J s
1.822 ×10−24 kg m/s
≈3.64 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 3.64 ×
10−10 m.
2. To determine whether the electron exhibits wave-like or particle-like be-
havior, we can compare the de Broglie wavelength λwith the size of the object.
25
If the de Broglie wavelength is comparable to or larger than the size of the
object, the wave-like behavior becomes significant.
In this case, the de Broglie wavelength of the electron is 3.64 ×10−10 m,
which is in the order of angstroms (roughly the size of an atom). Since the de
Broglie wavelength is comparable to the size of an atom, the electron exhibits
wave-like behavior.
Question 30
Question
An electron of mass 9.11 ×10−31 kg is moving with a velocity of 2.0×106m/s.
a) Calculate the de Broglie wavelength associated with the electron.
b) Discuss the implications of the de Broglie wavelength in the context of
the wave-particle duality.
Solution
a) To calculate the de Broglie wavelength associated with the electron, we use
the de Broglie wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34
J·s), and pis the momentum of the electron.
Step 1: Calculate the momentum pof the electron using the formula p=
mv.
p= (9.11 ×10−31 kg)(2.0×106m/s)
p= 1.82 ×10−24 kg ·m/s
Step 2: Now, substitute the values of hand pinto the de Broglie wavelength
formula:
λ=6.63 ×10−34 J·s
1.82 ×10−24 kg ·m/s
λ≈3.65 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 3.65 ×10−10 m.
b) The de Broglie wavelength is significant as it demonstrates the wave-
particle duality of matter. It suggests that particles, such as electrons, exhibit
both wave-like and particle-like properties. In the case of the electron, it can be
interpreted as having a wavelength associated with its motion, similar to a wave.
This duality challenges the classical view of particles as strictly localized entities
and highlights the need for a more complex understanding of the behavior of
subatomic particles.
26
Question 31
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
Solution
Step 1: Determine the kinetic energy of the electron. Since the electron is
accelerated through a potential difference of 100 V, the kinetic energy of the
electron can be calculated using the formula:
Kinetic energy (KE) = e·potential difference
where eis the elementary charge (1.6 x 10−19 C) and the potential difference is
100 V.
KE = (1.6×10−19 C) ×100 V = 1.6×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength of
a particle is given by:
λ=h
p
where h= Planck’s constant = 6.63 ×10−34 J s, p= momentum of the electron
=√2me·KE, and me= mass of the electron = 9.11 ×10−31 kg.
Substitute the known values into the formula:
p=p2·9.11 ×10−31 kg ·1.6×10−17 J = q2.91 ×10−30 kg m2s−2= 1.7×10−15 kg m/s
λ=6.63 ×10−34 J s
1.7×10−15 kg m/s = 3.9×10−19 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is 3.9×10−19 m.
Question 32
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after acceleration. (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)
27
Solution
Step 1: Calculate the kinetic energy gained by the electron from the potential
difference.
Given that the potential difference is 100 V, we can determine the kinetic
energy gained by the electron using the equation:
Kinetic energy = Charge ×Potential difference
Kinetic energy = (1.6×10−19 C) ×(100 V) = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using its kinetic energy.
The kinetic energy of the electron can also be expressed in terms of its
velocity (v):
Kinetic energy = 1
2×mass ×v2
Solving for v:
1.6×10−17 J = 1
2×9.11 ×10−31 kg ×v2
v2=2×1.6×10−17
9.11 ×10−31
v2≈3.53 ×109m2/s2
v≈1.88 ×104m/s
Step 3: Calculate the de Broglie wavelength using the formula:
de Broglie wavelength = h
p=h
mv
where his Planck’s constant (6.63×10−34 J s), mis the mass of the electron,
and vis the velocity of the electron.
Plugging in the values:
de Broglie wavelength = 6.63 ×10−34 J s
(9.11 ×10−31 kg) ×(1.88 ×104m/s)
de Broglie wavelength ≈7.3×10−10 m
28
Question 33
Question
An electron is accelerated by a potential difference of 600 V. Calculate the de
Broglie wavelength of the electron after acceleration. (The mass of an electron
is 9.11 ×10−31 kg and the elementary charge is 1.60 ×10−19 C).
Solution
Step 1: Calculate the kinetic energy of the electron using the equation:
Kinetic energy (KE) = e×Potential difference
Given that the potential difference is 600 V and the elementary charge is
1.60 ×10−19 C:
KE = (1.60 ×10−19 C)(600 V)
KE = 9.60 ×10−17 J
Step 2: Calculate the speed of the electron using the equation:
KE =1
2mv2
Given that the mass of an electron is 9.11 ×10−31 kg:
9.60 ×10−17 =1
2×(9.11 ×10−31)×v2
v2=2×9.60 ×10−17
9.11 ×10−31
v≈6.37 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron using the equa-
tion:
λ=h
mv
Where his Planck’s constant (6.63 ×10−34 J.s).
λ=6.63 ×10−34
(9.11 ×10−31)(6.37 ×106)
λ≈7.27 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 7.27 ×10−10 m.
29
Question 34
Question
An electron is moving with a velocity of 2.0×106m/s. Calculate the de Broglie
wavelength associated with the electron.
Solution
Step 1: Write down the de Broglie wavelength formula: The de Broglie wave-
length of a particle is given by:
λ=h
p
where: λ= de Broglie wavelength, h= Planck’s constant (6.63 ×10−34 J·s),
p= momentum of the particle.
Step 2: Calculate the momentum of the electron: The momentum of a
particle is given by:
p=m·v
where: m= mass of the electron (9.11 ×10−31 kg), v= velocity of the electron.
Given that v= 2.0×106m/s, we have:
p= (9.11 ×10−31 kg) ·(2.0×106m/s)
p= 1.822 ×10−24 kg ·m/s
Step 3: Calculate the de Broglie wavelength: Substitute the momentum
value into the de Broglie wavelength formula:
λ=6.63 ×10−34 J·s
1.822 ×10−24 kg ·m/s
λ≈3.64 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron moving at
2.0×106m/s is approximately 3.64 ×10−10 m.
Question 35
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron. (Use the elementary charge
as e= 1.6×10−19 C and the mass of an electron as me= 9.11 ×10−31 kg.)
30
Step 4: Calculate the de Broglie wavelength of the electron using the formula
λ=h
p.
Step 4: λ=6.63 ×10−34 Js
6.07 ×10−24 kg m/s = 1.09 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
1.09 ×10−10 m.
Question 2
Question
An electron with a kinetic energy of 200 eV is accelerated through a potential
difference of 50 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the velocity of the electron using the kinetic energy.
Kinetic energy (KE) = 1
2mv2
Given that the mass of an electron m= 9.11 ×10−31 kg and the kinetic energy
KE = 200 eV, where 1 eV = 1.6×10−19 J, we have:
200 ×1.6×10−19 =1
2×9.11 ×10−31 ×v2
v=r2×200 ×1.6×10−19
9.11 ×10−31
Step 2: Substitute the value of vinto the de Broglie wavelength formula.
λ=h
p=h
mv
where h= 6.63 ×10−34 J s is the Planck’s constant.
Step 3: Calculate the de Broglie wavelength. Substitute m,v, and hinto
the formula and solve for λ.
λ=6.63 ×10−34
9.11 ×10−31 ×v
Question 3
Question
A particle is moving with a velocity of 1.5×106m/s. Calculate the de Broglie
wavelength associated with this particle.
2
Solution
Step 1: Recall the de Broglie wavelength formula:
λ=h
p
where: λ= de Broglie wavelength (m), h= Planck’s constant (6.626×10−34 m2kg/s),
p= momentum of the particle (kg m/s).
Step 2: First, we need to determine the momentum of the particle using the
formula:
p=mv
where: m= mass of the particle, v= velocity of the particle.
Step 3: Let’s assume the particle is an electron with mass 9.11 ×10−31 kg.
Calculate the momentum using the mass and velocity given:
p= (9.11 ×10−31 kg) ×(1.5×106m/s)
Step 4: Calculate the momentum of the electron:
p= 1.365 ×10−24 kg m/s
Step 5: Now, substitute the momentum into the de Broglie wavelength for-
mula:
λ=6.626 ×10−34 m2kg/s
1.365 ×10−24 kg m/s
Step 6: Calculate the de Broglie wavelength:
λ= 4.853 ×10−10 m
Therefore, the de Broglie wavelength associated with the particle moving at
1.5×106m/s is 4.853 ×10−10 m.
Question 4
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
Given that the potential difference is 100 V, we can calculate the kinetic
energy of the electron using the equation K.E. =eV , where eis the charge of
an electron (1.6×10−19 C) and Vis the potential difference.
3
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
Step 2: Use the kinetic energy to find the de Broglie wavelength of the
electron.
The de Broglie wavelength of a particle is given by the equation λ=h
p,
where his the Planck constant (6.626 ×10−34 J·s) and pis the momentum of
the electron.
Since p=√2mK.E. for a non-relativistic particle, where mis the mass of
the electron (9.11 ×10−31 kg),
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J)
p≈3.027 ×10−24 kg ·m/s
Therefore, the de Broglie wavelength of the electron is:
λ=6.626 ×10−34 J·s
3.027 ×10−24 kg ·m/s
λ≈2.19 ×10−10 m
Question 5
Question
A particle with mass mand speed vexhibits wave-particle duality. Determine
its de Broglie wavelength λin terms of mand v.
Solution
To determine the de Broglie wavelength λof a particle with mass mand speed
v, we can 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 particle.
Step 1: First, we need to find the momentum pof the particle in terms of
its mass mand speed v. The momentum pis given by:
p=mv
4
Step 2: Now, substitute p=mv into the de Broglie wavelength formula:
λ=h
mv
So, the de Broglie wavelength λin terms of the mass mand speed vis
λ=h
mv .
Question 6
Question
A particle with mass mand velocity vis accelerated through a potential differ-
ence V. Determine the de Broglie wavelength associated with this particle.
Solution
Step 1: Determine the kinetic energy of the particle. The kinetic energy Kof
the particle can be calculated using the formula:
K=1
2mv2
Step 2: Relate the potential difference with the kinetic energy. The potential
energy Vcan be converted into the kinetic energy through the equation:
eV =K
where eis the elementary charge.
Step 3: Express the velocity in terms of potential difference. From steps
1 and 2, we can relate the velocity vto the potential difference Vusing the
equations:
K=1
2mv2
eV =K
Step 4: Calculate the de Broglie wavelength. The de Broglie wavelength λ
associated with the particle can be found using the equation:
λ=h
p
where his the Planck’s constant and pis the momentum of the particle.
Step 5: Express the momentum in terms of mass and velocity. The momen-
tum pof the particle can be related to its mass mand velocity vthrough the
equation:
p=mv
5
Step 6: Substitute the expression for momentum into the equation for de
Broglie wavelength. Substitute p=mv into the equation λ=h
pto get:
λ=h
mv
Step 7: Substitute the velocity expression in terms of potential difference into
the de Broglie wavelength equation. Substitute v=q2eV
minto the equation
λ=h
mv to get the de Broglie wavelength associated with the particle accelerated
through a potential difference V.
Question 7
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 (1.6 x 10−19 C) and Vis the potential
difference (100 V).
KE = (1.6×10−19 C) ×(100 V) = 1.6×10−17 J
Step 2: Next, we can find the momentum of the electron using the formula
p=√2me·KE, where meis the mass of the electron (9.11 x 10−31 kg).
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J) ≈3.49 ×10−24 kg m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
λ=h
p, where his the Planck constant (6.63 x 10−34 J s).
λ=6.63 ×10−34 J s
3.49 ×10−24 kg m/s ≈1.90 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.90 ×10−10 m.
Question 8
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
6
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =eV ,
where eis the charge of an electron (1.6×10−19 C) and Vis the potential
difference. Step 2: Calculate the velocity of the electron using the formula
KE =1
2mv2, where mis the mass of an electron (9.11 ×10−31 kg). Step 3:
Calculate the de Broglie wavelength using the formula λ=h
mv , where his the
Planck constant (6.63 ×10−34 J s).
Step 1: The kinetic energy KE of the electron is given by
KE =eV = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Using the kinetic energy formula 1
2mv2=KE, we have
1
2(9.11 ×10−31 kg)v2= 1.6×10−17 J
v2=2×1.6×10−17
9.11 ×10−31
v2= 3.5133 ×1013 m2/s2
v≈5.92 ×106m/s
Step 3: The de Broglie wavelength λis given by
λ=h
mv =6.63 ×10−34 J s
9.11 ×10−31 kg ×5.92 ×106m/s
λ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron acceler-
ated through a potential difference of 100 V is 1.22 ×10−10 m.
Question 9
Question
A beam of electrons with a kinetic energy of 500 eV strikes a crystal. If the
crystal diffracts the electrons at an angle of 30◦relative to the incident direction,
what is the de Broglie wavelength of the electrons?
Solution
Step 1: Calculate the speed of the electrons using their kinetic energy. Step 2:
Use the speed of the electrons to find their momentum. Step 3: Apply the de
Broglie wavelength formula to determine the wavelength of the electrons.
7
Step 1: Calculate the speed of the electrons. The kinetic energy of the
electrons is given by the formula:
KE =1
2mv2
Here, KE = 500 eV = 500 ×1.6×10−19 J(1eV=1.6×10−19 J), and we know
the mass of an electron m= 9.11 ×10−31 kg. Solving for v:
v=r2KE
m=r2×500 ×1.6×10−19
9.11 ×10−31
Step 2: Use the speed of the electrons to find their momentum. The mo-
mentum of an electron is given by:
p=mv
Substitute the values of mand vinto the above equation to find p.
Step 3: Apply the de Broglie wavelength formula to determine the wave-
length of the electrons. The de Broglie wavelength of the electrons is given
by:
λ=h
p
where h= 6.626 ×10−34 J s is the Planck constant. Substitute the calculated
value of pinto the de Broglie wavelength formula to find the wavelength of the
electrons.
Question 10
Question
An electron is accelerated through a potential difference of 100 V. Determine
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 formula:
K.E. =qV =eV
where qis the charge of the electron, Vis the potential difference, and eis the
elementary charge. Substituting the given values:
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
8
Step 2: Use the kinetic energy to find the de Broglie wavelength. The de
Broglie wavelength of a particle can be calculated using the formula:
λ=h
p
where his the Planck constant and pis the momentum of the electron. The
momentum of the electron can be calculated using the formula:
p=√2mK.E.
where mis the mass of the electron. Substituting the given values:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈3.42 ×10−24 kg m/s
Step 3: Substitute the momentum into the de Broglie wavelength formula.
λ=6.626 ×10−34 J s
3.42 ×10−24 kg m/s
λ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.94 ×
10−10 m.
Question 11
Question
An electron is accelerated through a potential difference of 1000 V. Calculate
the de Broglie wavelength of the electron after acceleration. The mass of an
electron is 9.11 ×10−31 kg and the elementary charge is 1.60 ×10−19 C.
Solution
Step 1: Calculate the kinetic energy of the electron. Step 2: Use the de Broglie
wavelength formula to find the wavelength of the electron.
Step 1: Calculate the kinetic energy of the electron.
The potential energy gained by the electron as it is accelerated through the
potential difference is equal to its kinetic energy. The kinetic energy of the
electron can be calculated using the formula:
KE =q·V
Where: KE = kinetic energy, q= charge of the electron, and V= potential
difference.
Substitute the values into the formula:
9
KE = (1.60 ×10−19 C)·(1000 V)
KE = 1.60 ×10−16 J
Therefore, the kinetic energy of the electron is 1.60 ×10−16 J.
Step 2: Use the de Broglie wavelength formula to find the wavelength of
the electron.
The de Broglie wavelength of an electron is given by the formula:
λ=h
p
Where: λ= de Broglie wavelength, h= Planck’s constant (6.63 ×10−34 J
s), p= momentum.
The momentum of the electron can be calculated using the formula:
p=√2mKE
Where: m= mass of the electron, and KE = kinetic energy.
Substitute the values into the formula:
p=p2×(9.11 ×10−31 kg)×(1.60 ×10−16 J)
p≈4.23 ×10−23 kg ·m/s
Now, substitute the momentum into the de Broglie wavelength formula:
λ=6.63 ×10−34 Js
4.23 ×10−23 kg ·m/s
λ≈1.57 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.57 ×10−10 m.
Question 12
Question
An electron is accelerated by a potential difference of 200 V. Calculate the de
Broglie wavelength of the electron.
10
Solution
Step 1: We first need to find the kinetic energy of the electron using the for-
mula K.E. =qV , where qis the charge of the electron and Vis the potential
difference.
K.E. =qV
= (1.6×10−19 C)(200 V)
= 3.2×10−17 J
Step 2: Next, we will use the equation for de Broglie wavelength, λ=h
p,
where his the Planck constant (6.626 ×10−34 J·s), pis the momentum of the
electron, and λis the de Broglie wavelength. We can find the momentum of the
electron using the formula p=√2mK.E., where mis the mass of the electron.
p=√2mK.E.
=p2×9.11 ×10−31 kg ×3.2×10−17 J
= 8.53 ×10−13 kg ·m/s
Step 3: Now, we can compute the de Broglie wavelength of the electron:
λ=h
p
=6.626 ×10−34 J·s
8.53 ×10−13 kg ·m/s
= 7.77 ×10−11 m
Therefore, the de Broglie wavelength of the electron accelerated by a poten-
tial difference of 200 V is 7.77 ×10−11 meters.
Question 13
Question
An electron is accelerated through a potential difference of 120 V. Calculate the
de Broglie wavelength associated with the electron.
(Note: The charge of an electron is −1.6×10−19 C and its mass is 9.11×10−31
kg.)
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =qV ,
where qis the charge and Vis the potential difference.
KE = (1.6×10−19 C)(120 V)
= 1.92 ×10−17 J
11
Step 2: Use the kinetic energy to calculate the speed of the electron using
the kinetic energy formula KE =1
2mv2, where mis the mass of the electron
and vis the speed.
1.92 ×10−17 =1
2(9.11 ×10−31)v2
v=r2×1.92 ×10−17
9.11 ×10−31
v≈5.94 ×106m/s
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 (p=mv).
λ=6.626 ×10−34 J s
9.11 ×10−31 kg ×5.94 ×106m/s
λ≈1.22 ×10−10 m
Therefore, the de Broglie wavelength associated with the accelerated electron
is approximately 1.22 ×10−10 m.
Question 14
Question
An electron with a kinetic energy of 500 eV exhibits wave-like behavior. Deter-
mine the de Broglie wavelength of the electron.
Solution
Step 1: Convert the kinetic energy of the electron from electron volts to joules.
Step 2: Use the de Broglie wavelength formula to calculate the wavelength of
the electron.
Step 1: To convert the kinetic energy from electron volts (eV) to joules (J),
we use the conversion factor 1 eV = 1.6×10−19 J:
Kinetic energy = 500 eV ×1.6×10−19 J/eV = 8 ×10−17 J
Step 2: The de Broglie wavelength (λ) of a particle with momentum pis
given by the formula:
λ=h
p
where his the Planck constant (6.63 ×10−34 J·s) and pis the momentum of
the particle. The momentum of an electron can be calculated using its kinetic
energy:
p=√2mE
12
where mis the mass of the electron (9.11×10−31 kg) and Eis the kinetic energy
(in joules).
Substitute the values into the formula:
p=p2×9.11 ×10−31 kg ×8×10−17 J≈5.35 ×10−24 kg m/s
Now, substitute hand pinto the de Broglie wavelength formula:
λ=6.63 ×10−34 J·s
5.35 ×10−24 kg m/s ≈1.24 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.24 ×
10−10 m.
Question 15
Question
An electron with kinetic energy of 200 eV is accelerated through a potential
difference of 50 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: First, we need to convert the kinetic energy of the electron from electron
volts to joules. We know that 1 eV is equivalent to 1.6×10−19 J. So, the kinetic
energy in joules is given by:
KE = 200 ×1.6×10−19 = 3.2×10−17 J
Step 2: Next, we can find the velocity of the electron using the kinetic energy
formula: KE =1
2mv2, where mis the mass of the electron and vis its velocity.
The mass of an electron is approximately 9.11 ×10−31 kg. Thus, we have:
3.2×10−17 =1
2×9.11 ×10−31 ×v2
v=r2×3.2×10−17
9.11 ×10−31
v≈5.36 ×106m/s
Step 3: Now, we can calculate the de Broglie wavelength of the electron
using its momentum, p=mv, and the de Broglie wavelength formula: λ=h
p,
where his the Planck constant. The momentum of the electron is given by:
p= 9.11 ×10−31 ×5.36 ×106= 4.88 ×10−24 kg m/s
Thus, the de Broglie wavelength is:
λ=6.626 ×10−34
4.88 ×10−24
13
λ≈1.36 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 1.36 ×
10−10 meters.
Question 16
Question
An electron is accelerated through a potential difference of 100 V. What is
the de Broglie wavelength of the electron after acceleration? Given that the
elementary charge of an electron is e= 1.6×10−19 C, the mass of an electron
is me= 9.11 ×10−31 kg, and Planck’s constant is h= 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron. The kinetic energy of an
electron accelerated through a potential difference ∆Vis given by the equation:
K.E. =e·∆V
Substitute the values for the elementary charge of an electron and the potential
difference into the equation:
K.E. = (1.6×10−19 C) ·(100 V) = 1.6×10−17 J
Step 2: Calculate the de Broglie wavelength using the kinetic energy. The de
Broglie wavelength of a particle with kinetic energy K.E. and mass mis given
by the equation:
λ=h
√2mK.E.
Substitute the values for Planck’s constant, the mass of an electron, and the
kinetic energy of the electron into the equation:
λ=6.63 ×10−34 J·s
p2·(9.11 ×10−31 kg) ·(1.6×10−17 J)
λ=6.63 ×10−34 J·s
p2·(9.11 ×10−14)
λ=6.63 ×10−34 J·s
√1.82 ×10−13
λ=6.63 ×10−34 J·s
1.35 ×10−6
λ= 4.89 ×10−28 m
Therefore, the de Broglie wavelength of the electron after acceleration is
4.89 ×10−28 m.
14
Question 17
Question
A photon with energy 3.0×10−19 J is incident on a metal surface. Calculate
the de Broglie wavelength of an electron ejected from the metal surface if the
work function of the metal is 4.2×10−19 J.
Solution
Step 1: Calculate the kinetic energy of the ejected electron. Given that the
energy of the incident photon is 3.0×10−19 J, and the work function of the
metal is 4.2×10−19 J, the kinetic energy (K.E.) of the ejected electron can be
calculated as:
K.E. =Ephoton −Work Function
K.E. = 3.0×10−19 −4.2×10−19
K.E. =−1.2×10−19 J
Step 2: Calculate the speed of the ejected electron using the kinetic energy.
The kinetic energy of the electron can be related to its speed using the equation
K.E. =1
2mv2, where mis the mass of an electron.
1
2mv2=−1.2×10−19
v2=−2× −1.2×10−19
9.1×10−31
v≈4.71 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength of an electron can be calculated using the formula λ=h
mv , where h
is Planck’s constant.
λ=h
mv =6.63 ×10−34
9.1×10−31 ×4.71 ×106
λ≈1.51 ×10−10 m
Therefore, the de Broglie wavelength of the ejected electron is approximately
1.51 ×10−10 m.
Question 18
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron.
15
Solution
Step 1: Calculate the kinetic energy of the electron using the formula K.E. =
qV , where qis the charge of the electron and Vis the potential difference.
Step 1: K.E. =qV = (1.6×10−19 C)(500 V) = 8 ×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron using the
formula K.E. =1
2mv2, where mis the mass of the electron.
Step 2: 1
2mv2= 8×10−17 J⇒v=r2K.E.
m=s2(8 ×10−17 J)
9.11 ×10−31 kg ≈1.49×106m/s
Step 3: 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.
Step 3: p=mv = (9.11 ×10−31 kg)(1.49 ×106m/s) ≈1.362 ×10−24 kg m/s
λ=h
p=6.63 ×10−34 J s
1.362 ×10−24 kg m/s ≈4.87 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 500 V is approximately 4.87 ×10−10 m.
Question 19
Question
A particle of mass mand velocity vexhibits wave-like behavior according to
wave-particle duality. Calculate the de Broglie wavelength of the particle in
terms of its momentum.
Solution
To determine the de Broglie wavelength of a particle in terms of its momentum,
we can use the de Broglie wavelength formula:
λ=h
p
where: - λis the de Broglie wavelength, - his the Planck constant (6.62607015×
10−34m2kg/s), - pis the momentum of the particle.
Step 1: Determine the momentum of the particle using its mass
and velocity The momentum of a particle is given by the formula:
p=m·v
where: - mis the mass of the particle, - vis the velocity of the particle.
16
Step 2: Substitute the momentum into the de Broglie wavelength
formula Substitute the expression for momentum (p=m·v) into the de Broglie
wavelength formula:
λ=h
m·v
Step 3: Simplify the expression The de Broglie wavelength can be sim-
plified to:
λ=h
m·v
Therefore, the de Broglie wavelength of the particle in terms of its momen-
tum is λ=h
m·v.
Question 20
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength of the electron after acceleration. (The charge of an
electron is −1.6×10−19 C and the mass of an electron is 9.11 ×10−31 kg.)
Solution
Step 1: Calculate the kinetic energy of the electron using the potential differ-
ence. The potential energy gained by the electron is equal to the kinetic energy
gained. Given that the charge of an electron is −1.6×10−19 C and the poten-
tial difference is 100 V, we can use the formula for potential energy: P E =qV .
Substitute the values in:
P E =−1.6×10−19 ×100 = −1.6×10−17 J
Since the potential energy gained equals the kinetic energy gained, the kinetic
energy of the electron is 1.6×10−17 J.
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy of the electron is given by:
KE =1
2mv2
where mis the mass of the electron. Substitute the values in and solve for v:
1.6×10−17 =1
2×9.11 ×10−31 ×v2
v≈5.352 ×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the electron.
The de Broglie wavelength of a particle is given by:
λ=h
mv
17
where his the Planck constant (6.63 ×10−34 J s). Substitute the values in:
λ=6.63 ×10−34
9.11 ×10−31 ×5.352 ×106
λ≈1.224 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.224 ×10−10 m.
Question 21
Question
An electron is accelerated through a potential difference of 200 V. Calculate the
de Broglie wavelength associated with the electron after acceleration.
Solution
We can use the de Broglie wavelength formula to find the wavelength associated
with 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 particle.
To find the momentum of the electron, we first need to find its kinetic energy
using the given potential difference. The kinetic energy of the electron is given
by:
KE =eV
where eis the charge of an electron (1.6×10−19 C), and Vis the potential
difference.
Step 1: Calculate the kinetic energy of the electron.
KE =eV = (1.6×10−19 C)(200 V)
KE = 3.2×10−17 J
Step 2: Calculate the momentum of the electron. The kinetic energy can
also be expressed in terms of the momentum p:
KE =p2
2m
where mis the mass of the electron (9.11 ×10−31 kg).
3.2×10−17 =p2
2(9.11 ×10−31)
18
p2= 6.4×10−17 ×2×9.11 ×10−31
p=p1.16384 ×10−30
p≈1.078 ×10−15
Step 3: Calculate the de Broglie wavelength.
λ=h
p=6.626 ×10−34
1.078 ×10−15
λ≈6.14 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 6.14 ×10−10 m.
Question 22
Question
An electron with a kinetic energy of 150 eV is incident on a single slit with a
width of 0.1 mm. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: We will first convert the given kinetic energy of the electron from electron
volts (eV) to joules (J). Given: E= 150 eV
Recall that 1 eV is equivalent to 1.602 ×10−19 J. Therefore, we have:
E= 150 ×1.602 ×10−19 J
Step 2: Now, we can calculate the velocity of the electron using the kinetic
energy formula:
E=1
2mv2
where mis the mass of the electron and vis the velocity. Since the mass of
an electron is 9.11 ×10−31 kg, we can solve for v.
Step 3: Next, we will calculate the de Broglie wavelength using the 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 electron.
Step 4: To find the momentum of the electron, we use the formula:
p=mv
Step 5: Finally, we substitute the calculated values of pand hinto the de
Broglie wavelength formula to find the wavelength of the electron.
19
Question 23
Question
An electron is accelerated from rest through a potential difference of 120 V. Cal-
culate the de Broglie wavelength associated with the electron after acceleration.
(Hint: The de Broglie wavelength λof a particle is given by λ=h
p, where h
is the Planck constant (6.626 ×10−34 m2kg/s) and pis the momentum of the
particle. For an electron, p=mv, where m= 9.11 ×10−31 kg is the mass of
the electron and vis the final velocity of the electron after acceleration. Use
the fact that the kinetic energy gained by the electron equals the work done on
it by the potential difference, i.e., eV =1
2mv2.)
Solution
Step 1: The kinetic energy gained by the electron is given by eV , where eis
the charge of an electron (1.6×10−19 C) and Vis the potential difference (120
V). This energy is equal to the kinetic energy of the electron after acceleration,
which is 1
2mv2. Setting eV =1
2mv2:
eV =1
2mv2
1.6×10−19 ×120 = 1
2×9.11 ×10−31 ×v2
1.92 ×10−17 = 4.55 ×10−31 ×v2
v2=1.92 ×10−17
4.55 ×10−31
v2≈4.23 ×1013 m/s
Step 2: The momentum pof the electron is equal to mv:
p=mv
p= 9.11 ×10−31 ×4.23 ×1013
p≈3.86 ×10−17 kg m/s
Step 3: Now, we can calculate the de Broglie wavelength λusing the formula
λ=h
p:
λ=h
p
λ=6.626 ×10−34
3.86 ×10−17
λ≈1.72 ×10−16 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 1.72 ×10−16 meters.
20
Question 24
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: First, we need to calculate the kinetic energy of the electron using the
formula K.E. =e·V, where eis the elementary charge and Vis the potential
difference. Given that the potential difference is 500 V, and the elementary
charge is 1.6×10−19 C, we have:
K.E. = 1.6×10−19 C×500 V
K.E. = 8 ×10−17 J
Step 2: Next, we can calculate the velocity of the electron using the for-
mula K.E. =1
2mv2, where mis the mass of the electron (9.11 x 10−31kg).8×
10−17 J = 1
2×9.11 ×10−31 kg ×v2
v=s2×8×10−17 J
9.11 ×10−31 kg
v= 714936.78 m/s
Step 3: Finally, we can calculate the de Broglie wavelength using the formula
λ=h
p, where his the Planck’s constant (6.63 x 10−34J.s)andp is the momentum
of the electron (m·v).
λ=6.63 ×10−34 J.s
9.11 ×10−31 kg ×714936.78 m/s
λ=6.63 ×10−34 J.s
6.51 ×10−24 kg.m/s
λ≈1.02 ×10−10 m
Therefore, the de Broglie wavelength associated with this electron is approx-
imately 1.02 ×10−10 m.
Question 25
Question
A proton is accelerated by a potential difference of 100 V. Calculate the de
Broglie wavelength of the proton using the formula λ=h
p, where λis the de
Broglie wavelength, his the Planck constant (6.63 ×10−34 m2kg/s), and pis
the momentum of the proton.
21
Solution
Step 1: Calculate the momentum of the proton using the formula p=√2mE,
where mis the mass of the proton and Eis the kinetic energy acquired by
the proton. Given that the charge of the proton, e= 1.6×10−19 C, and the
potential difference, V= 100 V, the kinetic energy can be calculated as:
E=eV = (1.6×10−19 C)(100 V)
Step 2: Substitute the kinetic energy into the formula p=√2mE to find
the momentum:
p=p2×(1.67 ×10−27 kg) ×E
Step 3: Calculate the de Broglie wavelength using the momentum obtained
in step 2:
λ=h
p
Question 26
Question
An electron is accelerated through a potential difference of 100 V. Calculate 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 his Planck’s constant (6.626 ×10−34 m2kg/s) and pis the momentum of
the electron.
Step 1: Find the kinetic energy of the electron The kinetic energy of
an electron accelerated through a potential difference ∆Vis given by:
K.E. =e·∆V
where eis the charge of an electron (1.6×10−19 C) and ∆Vis the potential
difference through which the electron is accelerated. Substituting the values,
we get:
K.E. = (1.6×10−19 C) ·(100 V) = 1.6×10−17 J
Step 2: Find the momentum of the electron The momentum of the
electron can be calculated using the formula:
p=√2mK.E.
22
where mis the mass of the electron (9.11 ×10−31 kg) and K.E. is the kinetic
energy of the electron. Substituting the values, we get:
p=p2×(9.11 ×10−31 kg) ×(1.6×10−17 J) ≈3.41 ×10−24 kg m/s
Step 3: Calculate the de Broglie wavelength Now, we can calculate
the de Broglie wavelength using the formula:
λ=h
p
Substitute the values of hand pto find the de Broglie wavelength:
λ=6.626 ×10−34 m2kg/s
3.41 ×10−24 kg m/s ≈1.94 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.94 ×10−10 m.
Question 27
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron.
Solution
Step 1: We can find the kinetic energy of the electron using the formula K.E. =
qV , where qis the charge of the electron and Vis the potential difference. Given
V= 100 V and charge of an electron q= 1.6×10−19 C,
K.E. = (1.6×10−19 C)(100 V)
K.E. = 1.6×10−17 J
Step 2: The de Broglie wavelength of a particle is given by the formula
λ=h
p, where his the Planck constant and pis the momentum. The momentum
of the electron can be calculated using p=√2mK.E., where mis the mass
of the electron and K.E. is the kinetic energy. Given mass of electron m=
9.11 ×10−31 kg,
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈3.4×10−24 kg m/s
Step 3: Now, we can find the de Broglie wavelength:
λ=h
p=6.63 ×10−34 J s
3.4×10−24 kg m/s
23
λ≈1.95 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.95 ×10−10 m.
Question 28
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Let’s start by using the kinetic energy equation for an electron accelerated
through a potential difference:
K=eV
where Kis the kinetic energy, eis the elementary charge (1.6×10−19 C), and
Vis the potential difference.
Step 1: Calculate the kinetic energy of the electron.
K=eV
K= (1.6×10−19 C)(100 V)
K= 1.6×10−17 J
Step 2: Use the de Broglie wavelength formula to find the wavelength
associated with the electron:
λ=h
p=h
√2mK
where his the Planck constant (6.63 ×10−34 J·s), mis the mass of the electron
(9.11 ×10−31 kg), and Kis the kinetic energy.
Step 3: Substitute the values into the formula to find the de Broglie wave-
length.
λ=h
√2mK
λ=6.63 ×10−34 J·s
p2×9.11 ×10−31 kg ×1.6×10−17 J
λ≈6.63 ×10−34
5.39 ×10−24 m
λ≈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 m.
24
Question 29
Question
An electron with a kinetic energy of 200 eV is traveling at a speed of 2 ×106
m/s. Determine:
1. The de Broglie wavelength of the electron.
2. Whether the electron exhibits wave-like or particle-like behavior.
Solution
1. To find the de Broglie wavelength of the electron, we can use the de Broglie
wavelength formula:
λ=h
p
where
p=mv
and mis the mass of the electron, vis its speed, his the Planck constant, and
λis the de Broglie wavelength.
Given:
Kinetic energy, K= 200 eV = 200 ×1.6×10−19 J
Speed, v= 2 ×106m/s
Mass of electron, m= 9.11 ×10−31 kg
Planck constant, h= 6.63 ×10−34 J s
First, calculate the momentum p:
p=mv
= (9.11 ×10−31 kg)(2 ×106m/s)
= 1.822 ×10−24 kg m/s
Now, calculate the de Broglie wavelength λ:
λ=h
p
=6.63 ×10−34 J s
1.822 ×10−24 kg m/s
≈3.64 ×10−10 m
Therefore, the de Broglie wavelength of the electron is approximately 3.64 ×
10−10 m.
2. To determine whether the electron exhibits wave-like or particle-like be-
havior, we can compare the de Broglie wavelength λwith the size of the object.
25
If the de Broglie wavelength is comparable to or larger than the size of the
object, the wave-like behavior becomes significant.
In this case, the de Broglie wavelength of the electron is 3.64 ×10−10 m,
which is in the order of angstroms (roughly the size of an atom). Since the de
Broglie wavelength is comparable to the size of an atom, the electron exhibits
wave-like behavior.
Question 30
Question
An electron of mass 9.11 ×10−31 kg is moving with a velocity of 2.0×106m/s.
a) Calculate the de Broglie wavelength associated with the electron.
b) Discuss the implications of the de Broglie wavelength in the context of
the wave-particle duality.
Solution
a) To calculate the de Broglie wavelength associated with the electron, we use
the de Broglie wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.63 ×10−34
J·s), and pis the momentum of the electron.
Step 1: Calculate the momentum pof the electron using the formula p=
mv.
p= (9.11 ×10−31 kg)(2.0×106m/s)
p= 1.82 ×10−24 kg ·m/s
Step 2: Now, substitute the values of hand pinto the de Broglie wavelength
formula:
λ=6.63 ×10−34 J·s
1.82 ×10−24 kg ·m/s
λ≈3.65 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 3.65 ×10−10 m.
b) The de Broglie wavelength is significant as it demonstrates the wave-
particle duality of matter. It suggests that particles, such as electrons, exhibit
both wave-like and particle-like properties. In the case of the electron, it can be
interpreted as having a wavelength associated with its motion, similar to a wave.
This duality challenges the classical view of particles as strictly localized entities
and highlights the need for a more complex understanding of the behavior of
subatomic particles.
26
Question 31
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
Solution
Step 1: Determine the kinetic energy of the electron. Since the electron is
accelerated through a potential difference of 100 V, the kinetic energy of the
electron can be calculated using the formula:
Kinetic energy (KE) = e·potential difference
where eis the elementary charge (1.6 x 10−19 C) and the potential difference is
100 V.
KE = (1.6×10−19 C) ×100 V = 1.6×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength of
a particle is given by:
λ=h
p
where h= Planck’s constant = 6.63 ×10−34 J s, p= momentum of the electron
=√2me·KE, and me= mass of the electron = 9.11 ×10−31 kg.
Substitute the known values into the formula:
p=p2·9.11 ×10−31 kg ·1.6×10−17 J = q2.91 ×10−30 kg m2s−2= 1.7×10−15 kg m/s
λ=6.63 ×10−34 J s
1.7×10−15 kg m/s = 3.9×10−19 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is 3.9×10−19 m.
Question 32
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after acceleration. (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)
27
Solution
Step 1: Calculate the kinetic energy gained by the electron from the potential
difference.
Given that the potential difference is 100 V, we can determine the kinetic
energy gained by the electron using the equation:
Kinetic energy = Charge ×Potential difference
Kinetic energy = (1.6×10−19 C) ×(100 V) = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using its kinetic energy.
The kinetic energy of the electron can also be expressed in terms of its
velocity (v):
Kinetic energy = 1
2×mass ×v2
Solving for v:
1.6×10−17 J = 1
2×9.11 ×10−31 kg ×v2
v2=2×1.6×10−17
9.11 ×10−31
v2≈3.53 ×109m2/s2
v≈1.88 ×104m/s
Step 3: Calculate the de Broglie wavelength using the formula:
de Broglie wavelength = h
p=h
mv
where his Planck’s constant (6.63×10−34 J s), mis the mass of the electron,
and vis the velocity of the electron.
Plugging in the values:
de Broglie wavelength = 6.63 ×10−34 J s
(9.11 ×10−31 kg) ×(1.88 ×104m/s)
de Broglie wavelength ≈7.3×10−10 m
28
Question 33
Question
An electron is accelerated by a potential difference of 600 V. Calculate the de
Broglie wavelength of the electron after acceleration. (The mass of an electron
is 9.11 ×10−31 kg and the elementary charge is 1.60 ×10−19 C).
Solution
Step 1: Calculate the kinetic energy of the electron using the equation:
Kinetic energy (KE) = e×Potential difference
Given that the potential difference is 600 V and the elementary charge is
1.60 ×10−19 C:
KE = (1.60 ×10−19 C)(600 V)
KE = 9.60 ×10−17 J
Step 2: Calculate the speed of the electron using the equation:
KE =1
2mv2
Given that the mass of an electron is 9.11 ×10−31 kg:
9.60 ×10−17 =1
2×(9.11 ×10−31)×v2
v2=2×9.60 ×10−17
9.11 ×10−31
v≈6.37 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron using the equa-
tion:
λ=h
mv
Where his Planck’s constant (6.63 ×10−34 J.s).
λ=6.63 ×10−34
(9.11 ×10−31)(6.37 ×106)
λ≈7.27 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 7.27 ×10−10 m.
29
Question 34
Question
An electron is moving with a velocity of 2.0×106m/s. Calculate the de Broglie
wavelength associated with the electron.
Solution
Step 1: Write down the de Broglie wavelength formula: The de Broglie wave-
length of a particle is given by:
λ=h
p
where: λ= de Broglie wavelength, h= Planck’s constant (6.63 ×10−34 J·s),
p= momentum of the particle.
Step 2: Calculate the momentum of the electron: The momentum of a
particle is given by:
p=m·v
where: m= mass of the electron (9.11 ×10−31 kg), v= velocity of the electron.
Given that v= 2.0×106m/s, we have:
p= (9.11 ×10−31 kg) ·(2.0×106m/s)
p= 1.822 ×10−24 kg ·m/s
Step 3: Calculate the de Broglie wavelength: Substitute the momentum
value into the de Broglie wavelength formula:
λ=6.63 ×10−34 J·s
1.822 ×10−24 kg ·m/s
λ≈3.64 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron moving at
2.0×106m/s is approximately 3.64 ×10−10 m.
Question 35
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron. (Use the elementary charge
as e= 1.6×10−19 C and the mass of an electron as me= 9.11 ×10−31 kg.)
30
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =eV ,
where Vis the potential difference.
KE = e·V= (1.6×10−19 C) ·(100 V) = 1.6×10−17 J
Step 2: Next, calculate the velocity of the electron using the formula for
kinetic energy: KE =1
2mv2, where mis the mass of the electron and vis the
velocity.
v=r2·KE
m=s2×1.6×10−17 J
9.11 ×10−31 kg ≈4.19 ×106m/s
Step 3: Now, calculate the de Broglie wavelength using the formula λ=h
p,
where his the Planck constant and pis the momentum.
λ=h
mv =6.63 ×10−34 J s
9.11 ×10−31 kg ×4.19 ×106m/s ≈1.70 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through 100 V potential difference is approximately 1.70 ×10−10 m.
31
