CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Wave-particle duality and de Broglie
wavelength
Question Bank - Set 1
Liberty University
Question 1
Question
An electron is accelerated through a potential difference of 120 V. What is the
de Broglie wavelength of the electron after being accelerated?
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.
Given V= 120 V, q = 1.6×10−19 C
KE = 1.6×10−19 ×120 = 1.92 ×10−17 J
Step 2: Use the formula for kinetic energy to find the momentum, p, of the
electron.
KE =p2
2m
Where mis the mass of the electron and pis its momentum.
m= 9.11 ×10−31 kg (mass of electron)
p=√2mKE =p2×9.11 ×10−31 ×1.92 ×10−17 = 2.94 ×10−24 kg m/s
Step 3: Calculate the de Broglie wavelength, λ, using the formula λ=h
p.
Where
h= 6.63 ×10−34 J s (Planck’s constant)
λ=6.63 ×10−34
2.94 ×10−24 = 2.25 ×10−10 m or 0.225 nm
Therefore, the de Broglie wavelength of the electron after being accelerated
through a potential difference of 120 V is 0.225 nm.
Question 2
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron.
Solution
Step 1: Determine the kinetic energy of the electron. We can use the rela-
tionship between potential energy (PE) and kinetic energy (KE) in an electron
accelerated through a potential difference:
PE = KE = eV
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(100 V). Substituting in the values:
KE = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Use the formula for kinetic energy in terms of de Broglie wavelength:
KE = 1
2mv2=h2
λ2m
where mis the mass of the electron (9.11 ×10−31 kg), vis the velocity of the
electron, λis the de Broglie wavelength, and his the Planck constant (6.63 ×
10−34 J·s).
Step 3: Rearrange the formula to solve for λ:
λ=h
√2mKE
Step 4: Substitute the values of h,m, and KE into the de Broglie wavelength
formula:
λ=6.63 ×10−34 J·s
p2(9.11 ×10−31 kg)(1.6×10−17 J)
Step 5: Calculate the de Broglie wavelength using the formula:
λ≈7.3×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 7.3×10−10 m.
2
Question 3
Question
An electron accelerated through a potential difference of 500 V has a kinetic
energy of 4.0 eV. Determine the de Broglie wavelength associated with this
electron.
Solution
Step 1: Given data: The potential difference (V) = 500 V Kinetic energy of the
electron (K.E.) = 4.0 eV
Step 2: Convert kinetic energy to joules: Since 1 eV = 1.6 x 10−19 J, the
kinetic energy in joules:
K.E. = 4.0 eV = 4.0×1.6×10−19 J=6.4×10−19 J
Step 3: Relation between kinetic energy and potential difference for an elec-
tron: The kinetic energy of an electron accelerated through a potential difference
(V) is given by:
K.E. =eV
where eis the charge of an electron (1.6 x 10−19 C).
Step 4: Solve for V:
K.E. =eV
V=K.E.
e=6.4×10−19
1.6×10−19 = 4 V
Step 5: Calculate the speed of the electron: The speed of the electron can
be calculated using the kinetic energy:
K.E. =1
2mv2
Solving for vgives:
v=r2K.E.
m
where mis the mass of the electron (9.11 x 10−31 kg).
Step 6: Substitute values and find the speed:
v=r2×6.4×10−19
9.11 ×10−31 = 8.19 ×106ms−1
Step 7: Calculate the de Broglie wavelength: The de Broglie wavelength (λ)
is given by:
λ=h
p
where his the Planck’s constant (6.626 x 10−34 Js) and pis the momentum of
the electron which is given by m×v.
3
Step 8: Substitute values and find the de Broglie wavelength:
λ=h
m×v=6.626 ×10−34
9.11 ×10−31 ×8.19 ×106= 8.02 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is 8.02 ×
10−11 m.
Question 4
Question
An electron is accelerated through a potential difference of 200 V. Calculate
the de Broglie wavelength associated with the electron after acceleration. (Take
the charge of an electron e= 1.6×10−19 C and the mass of an electron m=
9.11 ×10−31 kg.)
Solution
Step 1: Calculate the kinetic energy of the electron
Given potential difference, V= 200 V, and charge of an electron, e= 1.6×
10−19 C, the kinetic energy of the electron can be calculated by:
K.E. =eV
K.E. = (1.6×10−19 C)(200 V)
K.E. = 3.2×10−17 J
Step 2: Calculate the velocity of the electron
The kinetic energy can also be written in terms of the velocity of the electron
as:
K.E. =1
2mv2
Solving for v:
v=r2×K.E.
m
v=s2×3.2×10−17 J
9.11 ×10−31 kg
v≈6.64 ×106m/s
Step 3: Calculate the de Broglie wavelength
4
The de Broglie wavelength of the electron is given by:
λ=h
mv
where his the Planck constant (6.626 ×10−34 m2kg/s). Substituting the
values:
λ=6.626 ×10−34 m2kg/s
(9.11 ×10−31 kg)(6.64 ×106m/s)
λ≈7.29 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 7.29 ×10−10 m.
Question 5
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =eV ,
where e= 1.6×10−19 C is the charge of an electron and V= 500 V is the
potential difference.
KE =eV
= (1.6×10−19 C)(500 V)
= 8 ×10−17 J
Step 2: Use the formula for kinetic energy to calculate the momentum of the
electron: p=√2mKE, where m= 9.11 ×10−31 kg is the mass of the electron.
p=√2mKE
=p2(9.11 ×10−31 kg)(8 ×10−17 J)
≈4.3×10−24 kg m/s
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
h= 6.63 ×10−34 J s is the Planck’s constant.
λ=h
p
=6.63 ×10−34 J s
4.3×10−24 kg m/s
≈1.54 ×10−10 m
5
Therefore, the de Broglie wavelength of the electron is approximately 1.54 ×
10−10 m.
Question 6
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 (λ) associated with the electron accelerated
through a potential difference of 100 V, we can use the de Broglie wavelength
formula:
λ=h
p=h
√2mE
where: h= Planck’s constant (6.626 ×10−34 J·s), m= mass of the electron
(9.11 ×10−31 kg), E= kinetic energy of the electron (qV where qis the charge
of the electron and Vis the potential difference).
Given that q(charge of the electron) = 1.6×10−19 C and V= 100 V, we
can calculate the kinetic energy Eand then find the de Broglie wavelength λ.
Step 1: Calculate the kinetic energy of the electron
E=qV = (1.6×10−19 C)(100 V)
E= 1.6×10−17 J
Step 2: Find the de Broglie wavelength
λ=h
√2mE =6.626 ×10−34 J·s
p2(9.11 ×10−31 kg)(1.6×10−17 J)
λ=6.626 ×10−34
√2×9.11 ×1.6×10−48
λ=6.626 ×10−34
√29.056 ×10−48
λ=6.626 ×10−34
5.39 ×10−24
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 meters.
6
Question 7
Question
An electron is accelerated through a potential difference of 150 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Find the kinetic energy of the electron using the relation KE =qV ,
where qis the charge of an electron and Vis the potential difference. Step 2:
Use the kinetic energy to find the momentum of the electron using the equation
p=√2meKE, where meis the mass of an electron. Step 3: Calculate the de
Broglie wavelength using the relation λ=h
p, where his the Planck constant.
Step 1: The charge of an electron q= 1.6×10−19 C and the potential
difference V= 150 V.
KE =qV = (1.6×10−19 C)(150 V) = 2.4×10−17 J
Step 2: The mass of an electron me= 9.11×10−31 kg.
p=p2meKE =p2(9.11 ×10−31 kg)(2.4×10−17 J) ≈4.16 ×10−24 kg m/s
Step 3: The Planck constant h= 6.63×10−34 J·s.
λ=h
p=6.63 ×10−34 J·s
4.16 ×10−24 kg m/s ≈1.59 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 1.59 ×10−10 m.
Question 8
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron after acceleration. (Mass of an electron =
9.11 ×10−31 kg, Charge of an electron = 1.60 ×10−19 C, Planck’s constant =
6.63 ×10−34 Js)
Solution
Step 1: Calculate the kinetic energy of the electron using the relation KE =qV ,
where KE is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Step 1: KE = (1.60 ×10−19 C)(500 V) = 8.00 ×10−17 J
7
Step 2: Calculate the velocity of the electron using the formula KE =1
2mv2,
where mis the mass of the electron and vis the velocity of the electron.
KE =1
2mv2
8.00 ×10−17 =1
2(9.11 ×10−31)v2
v=r2×8.00 ×10−17
9.11 ×10−31
v≈6.61 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron using the de
Broglie wavelength formula λ=h
p, where his Planck’s constant and pis the
momentum of the electron.
Step 3: p=mv = (9.11 ×10−31 kg)(6.61 ×106m/s)
p= 6.02 ×10−24 kg m/s
λ=6.63 ×10−34 J s
6.02 ×10−24 kg m/s
λ≈1.10 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 500 V is approximately 1.10 ×10−10 m.
Question 9
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: We can use the de Broglie wavelength formula, which relates the mo-
mentum of a particle to its wavelength:
λ=h
p
where: λ= de Broglie wavelength (m), h= Planck’s constant (6.626×10−34 J·s),
p= momentum of the particle (kg m/s).
Step 2: The momentum of the electron can be calculated using its kinetic
energy. The kinetic energy of the electron can be expressed as:
K.E. =1
2mv2=eV
8
where: K.E. = kinetic energy (J), m= mass of electron (9.11 ×10−31 kg), v=
velocity of electron (m/s), e= elementary charge (1.60×10−19 C), V= potential
difference (V).
Step 3: We can rearrange the kinetic energy equation to solve for the velocity
of the electron:
v=r2eV
m
Step 4: Substituting the known values into the equation for velocity:
v=s2×1.60 ×10−19 C×100 V
9.11 ×10−31 kg
Step 5: Calculating the velocity of the electron:
v≈6.58 ×106m/s
Step 6: Now, we can calculate the momentum of the electron using the
formula p=mv.
Step 7: Substituting the mass and velocity of the electron into the momen-
tum equation:
p= (9.11 ×10−31 kg)(6.58 ×106m/s)
Step 8: Calculating the momentum of the electron:
p≈6.00 ×10−24 kg m/s
Step 9: Finally, substitute the momentum into the de Broglie wavelength
formula:
λ=(6.626 ×10−34 J·s)
6.00 ×10−24 kg m/s
Step 10: Calculating the de Broglie wavelength of the electron:
λ≈1.11 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.11 ×10−10 m.
Question 10
Question
An electron with a kinetic energy of 3.0×10−19 J, moving in a vacuum, exhibits
wave-like behavior. Calculate the de Broglie wavelength associated with this
electron.
9
Solution
Step 1: Recall the de Broglie wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34
Js), and pis the momentum of the particle.
Step 2: We can find the momentum of the electron using the formula:
K.E. =1
2mv2=p2
2m
where mis the mass of the electron and vis its velocity.
Step 3: Since v=q2K.E.
m, we can substitute this into the momentum
formula to get:
p=√2mK.E.
Step 4: Now we substitute pinto the de Broglie wavelength formula:
λ=h
√2mK.E.
Step 5: Plugging in the known values, we have:
λ=6.626 ×10−34 Js
p2×9.11 ×10−31 kg ×3.0×10−19 J
Step 6: Solving this expression gives:
λ=6.626 ×10−34 Js
p2×9.11 ×10−31 kg ×3.0×10−19 J= 2.74 ×10−12 m
Therefore, the de Broglie wavelength associated with the electron is 2.74 ×
10−12 m.
Question 11
Question
An electron is accelerated through a potential difference of 50 V. Determine the
de Broglie wavelength associated with the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential differ-
ence. Step 2: Utilize the de Broglie wavelength formula to find the wavelength
associated with the electron.
10
Step 1: Calculate the kinetic energy of the electron using the potential
difference.
The kinetic energy of an electron accelerated through a potential difference
can be calculated using the equation:
K.E. =eV
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(50 V).
Substitute the values to find the kinetic energy:
K.E. = (1.6×10−19 C) ×(50 V) = 8 ×10−18 J
Step 2: Utilize the de Broglie wavelength formula to find the wavelength
associated with the electron.
The de Broglie wavelength (λ) of a particle with kinetic energy K.E. and
mass mcan be calculated using the formula:
λ=h
p=h
√2mK.E.
where his the Planck constant (6.626 ×10−34 m2kg/s) and mis the mass of the
electron (9.109 ×10−31 kg).
Substitute the values into the formula to find the de Broglie wavelength:
λ=6.626 ×10−34 m2kg/s
p2×9.109 ×10−31 kg ×8×10−18 J
λ=6.626 ×10−34 m2kg/s
√1.451 ×10−11
λ≈6.626 ×10−34 m2kg/s
1.205 ×10−5
λ≈5.496 ×10−30 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 50 V is approximately 5.496 ×10−30 m.
Question 12
Question
An electron is accelerated through a potential difference of 1000 V. Determine
the de Broglie wavelength of the electron.
11
Solution
Step 1: The kinetic energy of the electron can be calculated using the formula:
KE =eV
where eis the charge of an electron (1.6 x 10−19 C), and Vis the potential
difference (1000 V in this case). Therefore,
KE = (1.6×10−19 C)(1000 V)
KE = 1.6×10−16 J
Step 2: The velocity of the electron can be determined using the kinetic
energy formula:
KE =1
2mv2
where mis the mass of an electron (9.11 x 10−31 kg). Therefore,
1.6×10−16 =1
2(9.11 ×10−31)v2
v2=2(1.6×10−16)
9.11 ×10−31
v≈3.53 ×106m/s
Step 3: The de Broglie wavelength of the electron can be calculated using
the formula:
λ=h
p=h
mv
where his the Planck constant (6.63 ×10−34 J·s). Substituting the values,
λ=6.63 ×10−34
(9.11 ×10−31)(3.53 ×106)
λ≈2.47 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 1000 V is approximately 2.47 ×10−10 m.
Question 13
Question
An electron is accelerated through a potential difference of 200 V. Determine
the de Broglie wavelength associated with the electron’s motion.
12
Solution
Step 1: Recall 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 particle.
Step 2: Next, we need to find the momentum of the electron. The kinetic
energy gained by the electron is given by:
K.E. =qV
where qis the charge of the electron (1.6×10−19 C) and Vis the potential
difference (200 V).
Step 3: The kinetic energy can also be expressed in terms of momentum as:
K.E. =p2
2m
where mis the mass of the electron (9.11 ×10−31 kg).
Step 4: Equating the two expressions for kinetic energy, we have:
qV =p2
2m
Step 5: Solving for p, we get:
p=p2m·qV
Step 6: Substitute the known values to find p:
p=p2·9.11 ×10−31 kg ·1.6×10−19 C·200 V
Step 7: Calculate pto get:
p≈1.48 ×10−24 kg ·m/s
Step 8: Finally, substitute pinto the de Broglie wavelength formula to find
λ:
λ=6.63 ×10−34 J·s
1.48 ×10−24 kg ·m/s
Step 9: Calculate λto get the de Broglie wavelength associated with the
electron’s motion.
Question 14
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
13
Solution
Step 1: Start with the formula for de Broglie wavelength:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J
s), and pis the momentum of the particle.
Step 2: Calculate the momentum of the electron from its kinetic energy
gained by passing through the potential difference:
K.E. =e×V=1
2mv2
where eis the charge of an electron (1.6×10−19 C), Vis the potential difference
(100 V), mis the mass of an electron (9.11 ×10−31 kg), and vis the velocity of
the electron.
Step 3: The velocity of an electron with kinetic energy K.E. can be expressed
as:
v=r2×K.E.
m
Step 4: Substitute the given values and solve for v:
v=r2×(e×V)
m
Step 5: Once you have the velocity, you can calculate the momentum pusing
p=mv.
Step 6: Finally, substitute the momentum pinto the de Broglie wavelength
formula to find the wavelength λ. Remember to convert the momentum to SI
units (kg m/s) if necessary.
Question 15
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 electron, e= 1.6×10−19 C Mass of electron, m= 9.11 ×
10−31 kg Planck’s constant, h= 6.63 ×10−34 J·s Speed of light, c= 3.00 ×108
m/s
Solution
Step 1: Calculate the kinetic energy of the electron after acceleration. Given
that the potential difference is 100 V, the kinetic energy can be calculated using
14
the formula: KE =qV , where qis the charge and Vis the potential difference.
Calculating the value of KE:
KE = 1.6×10−19 C×100 V = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using the kinetic energy. The
kinetic energy can be expressed as:
KE =1
2mv2
where mis the mass of the electron and vis the velocity. Solving for v:
v=r2KE
m=s2×1.6×10−17 J
9.11 ×10−31 kg ≈3.93 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength is given by the formula:
λ=h
mv
Substitute the values of h,m, and vinto the formula:
λ=6.63 ×10−34 J·s
9.11 ×10−31 kg ×3.93 ×106m/s
λ≈1.70 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.70 ×10−10 m.
Question 16
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
given. Step 2: Use the kinetic energy to find the momentum of the electron.
Step 3: Apply de Broglie’s wavelength formula to determine the de Broglie
wavelength.
Step 1: The kinetic energy of the electron can be calculated using the
formula:
Kinetic energy (KE) = q·potential difference
15
where qis the charge of the electron. The charge of an electron is −1.6×10−19
C. Substituting the values:
KE = (−1.6×10−19 C) ×(100 V)
KE = −1.6×10−17 J
Step 2: The kinetic energy of the electron can also be expressed in terms
of momentum pas:
KE = p2
2m
where mis the mass of the electron. The mass of an electron is 9.11 ×10−31
kg. Rearranging the equation to solve for momentum:
p=√2m·KE = p2×9.11 ×10−31 kg ×1.6×10−17 J
p=√2×9.11 ×1.6×10−17
p≈8.55 ×10−17 kg m/s
Step 3: De Broglie’s wavelength formula is given by:
λ=h
p
where his the Planck constant (6.63 ×10−34 J s). Substituting the values:
λ=6.63 ×10−34 J s
8.55 ×10−17 kg m/s
λ≈7.77 ×10−11 m
Therefore, the de Broglie wavelength of the electron is approximately 7.77 ×
10−11 m.
Question 17
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron after acceleration. (Mass of an electron =
9.11 ×10−31 kg, charge of an electron = −1.6×10−19 C)
16
Solution
Step 1: First, we calculate the kinetic energy of the electron using the formula
KE =qV , where qis the charge and Vis the potential difference.
Step 1: KE =qV = (−1.6×10−19 C)(100 V) = −1.6×10−17 J
Step 2: Next, we calculate the velocity of the electron using the formula for
kinetic energy: KE =1
2mv2, where mis the mass of the electron.
Step 2: KE =1
2mv2=⇒v=r2·KE
m=s2·1.6×10−17 J
9.11 ×10−31 kg ≈2.44×106m/s
Step 3: Now, we can calculate the de Broglie wavelength of the electron using
the formula: λ=h
p, where his the Planck constant and pis the momentum.
Step 3: p=mv = (9.11 ×10−31 kg)(2.44 ×106m/s) ≈2.22 ×10−24 kg m/s
λ=h
p=6.63 ×10−34 J s
2.22 ×10−24 kg m/s ≈2.99 ×10−9m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 2.99 ×10−9m.
Question 18
Question
A beam of electrons is accelerated through a potential difference of 200 V.
Calculate the de Broglie wavelength of these electrons.
Solution
Step 1: Determine the kinetic energy of the electrons using the potential differ-
ence.
The kinetic energy of an electron accelerated through a potential difference
is given by the formula:
Ekin =qV
where qis the charge of an electron (−1.6×10−19 C) and Vis the potential
difference (200 V).
Ekin = (−1.6×10−19 C)(200 V)
Ekin =−3.2×10−17 J
Step 2: Calculate the momentum of the electrons.
17
The momentum of an electron can be calculated using the formula:
p=p2mEkin
where mis the mass of an electron (9.11 ×10−31 kg) and Ekin is the kinetic
energy calculated in Step 1.
p=p2(9.11 ×10−31 kg)(−3.2×10−17 J)
p≈1.32 ×10−24 kg m/s
Step 3: Find the de Broglie wavelength of the electrons using the formula:
λ=h
p
where his the Planck constant (6.626 ×10−34 J s) and pis the momentum
calculated in Step 2.
λ=6.626 ×10−34 J s
1.32 ×10−24 kg m/s
λ≈5.02 ×10−10 m
Therefore, the de Broglie wavelength of the electrons accelerated through a
potential difference of 200 V is approximately 5.02 ×10−10 m.
Question 19
Question
An electron with a kinetic energy of 200 eV is moving through space. Determine
the de Broglie wavelength associated with the electron.
Solution
Step 1: Convert the kinetic energy of the electron from electronvolts (eV) to
joules (J). Step 2: Use the de Broglie wavelength formula to find the wavelength
associated with the electron.
Step 1: Given: Kinetic energy of the electron = 200 eV
We know that 1 electronvolt is equal to 1.6×10−19 J. Therefore, the kinetic
energy of the electron in joules is:
200 eV ×1.6×10−19 J/eV = 3.2×10−17 J
Step 2: The de Broglie wavelength of a particle is given by:
λ=h
p
18
where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J s),
and pis the momentum of the particle.
The momentum of the electron can be calculated using the equation:
p=√2mK
where mis the mass of the electron (9.11×10−31 kg) and Kis the kinetic energy.
Substitute the values into the equation to find the momentum of the electron:
p=p2×9.11 ×10−31 kg ×3.2×10−17 J≈7.57 ×10−24 kg m/s
Now substitute the momentum into the de Broglie wavelength formula to
find the wavelength:
λ=6.626 ×10−34 J s
7.57 ×10−24 kg m/s ≈8.75 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is 8.75 ×
10−11 meters.
Question 20
Question
An electron is accelerated through a potential difference of 100 volts. Calculate
the de Broglie wavelength of the electron.
Solution
Step 1: We can start by finding the kinetic energy of the electron using the
formula K=eV , where eis the elementary charge (1.6×10−19 coulombs) and
Vis the potential difference (100 volts). Step 2: Substitute the values into the
formula to find the kinetic energy K. Step 3: Next, we can use the de Broglie
wavelength formula λ=h
p, where his the Planck constant (6.63×10−34 J·s) and
pis the momentum of the electron. Step 4: The momentum pcan be calculated
using the equation p=√2mK, where mis the mass of the electron (9.11×10−31
kg) and Kis the kinetic energy. Step 5: Substitute the kinetic energy Kinto
the momentum equation to find the momentum p. Step 6: Finally, substitute
the momentum pinto the de Broglie wavelength formula to calculate the de
Broglie wavelength λ.
Question 21
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
19
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 mis
the mass of the electron. Step 3: Use the de Broglie wavelength formula λ=h
p,
where his the Planck constant and pis the momentum of the electron. Step
4: Calculate the momentum of the electron using the formula p=mv. Step
5: Substitute the momentum of the electron into the de Broglie wavelength
formula to find the de Broglie wavelength.
Step 1: Calculate the kinetic energy of the electron. The elementary charge
e= 1.6×10−19 C. The potential difference V= 100 V. The kinetic energy
K=eV = (1.6×10−19 C)(100 V) = 1.6×10−17 J.
Step 2: Calculate the velocity of the electron. The mass of the electron
m= 9.11 ×10−31 kg. Using the kinetic energy formula: K=1
2mv21.6×
10−17 =1
2(9.11 ×10−31)(v2)v2=2(1.6×10−17 )
9.11×10−31 v2≈3.53 ×106v≈√3.53 ×106
v≈1.88 ×103m/s.
Step 3: Use the de Broglie wavelength formula. The Planck constant
h= 6.63 ×10−34 J s. The momentum of the electron p=mv = (9.11 ×
10−31 kg)(1.88 ×103m/s) = 1.71 ×10−27 kg m/s.
Step 4: Calculate the de Broglie wavelength. λ=h
p=6.63×10−34 J s
1.71×10−27 kg m/s
λ≈3.88 ×10−10 m.
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 3.88 ×10−10 m.
Question 22
Question
An electron is accelerated through a potential difference of 120 V. Determine
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 ×10−19 C) and Vis the potential
difference (120 V).
KE = (1.6×10−19 C) ×(120 V)
Step 2: Calculating the kinetic energy:
KE = 1.92 ×10−17 J
Step 3: Next, we can relate the kinetic energy of the electron to its de Broglie
wavelength using the formula:
KE =1
2mv2=h2
λ2·2m
20
Step 4: Rearranging the equation to solve for λ:
λ=h
√2mKE
Step 5: Plugging in the values of Planck’s constant h(6.626 ×10−34 J·s)
and the mass of an electron m(9.11 ×10−31 kg), and the calculated kinetic
energy:
λ=6.626 ×10−34 J·s
p2×9.11 ×10−31 kg ×1.92 ×10−17 J
Step 6: Solving for λ:
λ≈2.3×10−12 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 120 V is approximately 2.3×10−12 m.
Question 23
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after acceleration. (Take the electron
charge as 1.6×10−19 C and its mass as 9.11 ×10−31 kg)
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.
Given: q= 1.6×10−19 C, V = 100 V
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using the formula for kinetic
energy: KE =1
2mv2, where mis the mass of the electron.
Given: m= 9.11 ×10−31 kg
1.6×10−17 =1
2(9.11 ×10−31)v2
v2=2×1.6×10−17
9.11 ×10−31
v2=3.2×10−17
9.11 ×10−31
21
v2≈3.51 ×1013
v≈p3.51 ×1013
v≈5.92 ×106m/s
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
his the Planck’s constant and pis the momentum of the electron.
Given: h= 6.63 ×10−34 J s
p=m·v= (9.11 ×10−31 kg)(5.92 ×106m/s)
p≈5.40 ×10−24 kg m/s
λ=6.63 ×10−34 J s
5.40 ×10−24 kg m/s
λ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.23 ×10−10 m.
Question 24
Question
An electron with a mass of 9.11 ×10−31 kg is accelerated through a potential
difference of 80 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =qV ,
where qis the charge of the electron (we’ll assume it is −1.6×10−19 C) and V
is the potential difference.
KE = (1.6×10−19 C)(80 V)
KE = 1.28 ×10−17 J
Step 2: Use the kinetic energy to find the speed of the electron using the
formula KE =1
2mv2, where mis the mass of the electron.
1.28 ×10−17 =1
2×(9.11 ×10−31)v2
v2=2×1.28 ×10−17
9.11 ×10−31
v≈3.04 ×106m/s
22
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
his Planck’s constant and pis the momentum. The momentum can be calcu-
lated as mv.
p= (9.11 ×10−31)(3.04 ×106)
p≈2.77 ×10−24 kg m/s
λ=6.626 ×10−34
2.77 ×10−24
λ≈2.39 ×10−10 m
Therefore, the de Broglie wavelength of the electron is 2.39 ×10−10 m.
Question 25
Question
An electron is accelerated through a potential difference of 250 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Use the energy gained by the electron to find its kinetic energy. Step 2:
Apply the de Broglie wavelength formula to calculate the wavelength.
Step 1: Calculate the kinetic energy of the electron. The energy gained
by the electron is equal to the potential energy difference it was accelerated
through, which can be converted to kinetic energy. The kinetic energy is given
by the formula:
K=qV
where qis the charge of the electron (1.6 x 10-19 C) and Vis the potential
difference (250 V).
K= (1.6×10−19 C) ×(250 V)
K= 4 ×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength is
given by:
λ=h
p
where: h= Planck’s constant = 6.626 x 10-34 J
·
s, p= momentum of the electron,
and λ= de Broglie wavelength.
The momentum of the electron can be calculated using the kinetic energy:
p=√2mK
where m= mass of the electron = 9.11 x 10-31 kg and K= kinetic energy.
23
Substitute the values to find p:
p=p2×9.11 ×10−31 kg ×4×10−17 J
p≈1.19 ×10−23 kg m/s
Now, substitute hand pinto the de Broglie wavelength formula:
λ=6.626 ×10−34 J
·
s
1.19 ×10−23 kg m/s
λ≈5.57 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 5.57 ×10−11 meters.
Question 26
Question
A particle of mass mmoves with velocity vand has kinetic energy E. Determine
the de Broglie wavelength of this particle.
Solution
The de Broglie wavelength of a particle is given by the formula:
λ=h
p
where his the Planck constant (6.62607015 ×10−34 Js) and pis the momentum
of the particle.
Step 1: First, we need to express the momentum of the particle in terms of
its kinetic energy and mass. The momentum of the particle is given by:
p=mv
Step 2: Next, we can express the velocity vin terms of kinetic energy E
and mass m. The kinetic energy of the particle is given by:
E=1
2mv2
Solving for vgives:
v=r2E
m
Step 3: Substituting vback into the momentum formula, we get:
p=mr2E
m=√2mE
24
Step 4: Now, we can substitute the momentum pinto the formula for de
Broglie wavelength:
λ=h
√2mE =h
q2m·1
2mv2
=h
√mE
Hence, the de Broglie wavelength of the particle is h
√mE .
Question 27
Question
An electron is accelerated from rest through a potential difference of 1000 V.
Calculate the de Broglie wavelength of the electron after acceleration.
Solution
Step 1: Find the kinetic energy of the electron using the potential difference.
The kinetic energy of the electron can be calculated using the formula:
K.E. =e·V
where eis the elementary charge and Vis the potential difference. Given that
the potential difference is 1000 V, and the elementary charge e= 1.6×10−19
C, we have:
K.E. = 1.6×10−19 ×1000
K.E. = 1.6×10−16 J
Step 2: Use the kinetic energy to find the de Broglie wavelength. The de
Broglie wavelength of a particle is given by the formula:
λ=h
p
where his the Planck constant (6.626 ×10−34 m2kg/s) and pis the momentum
of the particle. The momentum of the electron can be calculated using the
kinetic energy:
p=√2mK.E.
where mis the mass of the electron (9.11 ×10−31 kg) and K.E. is the kinetic
energy. Substitute the known values into the momentum formula:
p=p2×9.11 ×10−31 ×1.6×10−16
p≈2.63 ×10−24 kg m/s
Now, substitute the momentum into the de Broglie wavelength formula:
λ=6.626 ×10−34
2.63 ×10−24
25
λ≈2.51 ×10−12 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 2.51 ×10−12 meters.
Question 28
Question
A neutron is accelerated through a potential difference of 50 V. Calculate the
de Broglie wavelength associated with the neutron after being accelerated. The
mass of a neutron is 1.67 ×10−27 kg and the charge of an electron is 1.6×10−19
C.
Solution
Step 1: Calculate the kinetic energy gained by the neutron. Given that the
potential difference is 50 V, the kinetic energy gained by the neutron can be
calculated using the equation:
K=qV
where: K= kinetic energy gained (J), q= charge of the particle (C), V=
potential difference (V).
Substitute the values:
K= (1.6×10−19C)×(50V)=8×10−18J
Step 2: Calculate the velocity of the neutron. The kinetic energy gained is
equal to the energy associated with the motion of the neutron, so we have:
K=1
2mv2
Solve for velocity (v):
v=r2K
m
v=s2×8×10−18J
1.67 ×10−27kg
v=r1.6×10−17
1.67 ×1010
v≈2.39 ×105m/s
Step 3: Calculate the de Broglie wavelength associated with the neutron.
The de Broglie wavelength (λ) is given by:
λ=h
p
26
where: h= Planck’s constant (6.63×10−34 J·s), p= momentum of the neutron
(m×v).
Substitute the values for momentum:
p= 1.67 ×10−27 kg ×2.39 ×105m/s
p= 4.0×10−22 kg m/s
Now, calculate the de Broglie wavelength:
λ=6.63 ×10−34 J·s
4.0×10−22 kg m/s
λ=6.63
4.0×10−12 m
λ≈1.66 ×10−12 m
Therefore, the de Broglie wavelength associated with the neutron after being
accelerated is approximately 1.66 ×10−12 m.
Question 29
Question
Consider an electron with a kinetic energy of 100 eV. If the electron behaves as
a wave, what is 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 J
s), and pis the momentum of the electron.
First, we need to find the momentum of the electron using the kinetic energy.
Given that the kinetic energy of the electron is 100 eV, we need to convert this
energy to joules:
1 eV = 1.602 ×10−19 joules
Thus, the kinetic energy in joules is:
100 eV ×1.602 ×10−19 J/eV = 1.602 ×10−17 J
Next, we can use the formula for kinetic energy to find the momentum p:
KE =p2
2m
27
where mis the mass of an electron (9.11 ×10−31 kg).
Step 1: Solve for momentum pin the kinetic energy formula:
p=√2mKE
p=p2×9.11 ×10−31 ×1.602 ×10−17
p=p2.9×10−32
p≈5.39 ×10−16 kg m/s
Now that we have the momentum of the electron, we can find the de Broglie
wavelength:
Step 2: Calculate the de Broglie wavelength:
λ=h
p
λ=6.626 ×10−34
5.39 ×10−16
λ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 1.23 ×10−10 meters.
Question 30
Question
A particle of mass mis confined to move in one dimension within a box of
length L. The energy of the particle is given by E=n2ℏ2π2
2mL2, where nis a
positive integer. Determine the de Broglie wavelength of the particle in terms
of Land n.
Solution
Step 1: The de Broglie wavelength of a particle is given by λ=h
p, where his
the Planck constant and pis the momentum of the particle.
Step 2: The momentum of the particle can be expressed in terms of its
energy Eusing the relation E=p2
2m. Rearranging, we have p=√2mE.
Step 3: Substitute the given expression for energy into the momentum equa-
tion to find the momentum in terms of n,ℏ,m, and L.
p=r2m·n2ℏ2π2
2mL2=nℏπ
L
28
Step 4: Now, substitute the expression for momentum into the de Broglie
wavelength formula to find the de Broglie wavelength λ.
λ=h
p=h
nℏπ/L =hL
nℏπ
Step 5: Simplify the expression to obtain the final result for the de Broglie
wavelength in terms of Land n.
λ=L
nπ
Question 31
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron in meters. Given that the charge of an
electron is −1.6×10−19 C and Planck’s constant is 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
The kinetic energy of the electron can be calculated using the formula:
K.E. =qV
where K.E. is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Substitute q=−1.6×10−19 C and V= 100 V into the formula:
K.E. =−1.6×10−19 C×100 V
Step 2: Calculate the velocity of the electron. The kinetic energy of an
electron can be expressed in terms of its velocity using the formula:
K.E. =1
2mv2
where K.E. is the kinetic energy, mis the mass of the electron, and vis the
velocity.
Equating the two expressions for kinetic energy:
−1.6×10−19 ×100 = 1
2×9.11 ×10−31 ×v2
Step 3: Solve for the velocity of the electron.
v=r−1.6×10−19 ×100 ×2
9.11 ×10−31
29
Step 4: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength of an electron can be calculated using the formula:
λ=h
p
where λis the de Broglie wavelength, his Planck’s constant, and pis the
momentum of the electron.
The momentum of the electron can be expressed as:
p=mv
Substitute m= 9.11 ×10−31 kg, vcalculated in step 3, and h= 6.63 ×10−34
J·s into the formula:
λ=6.63 ×10−34
9.11 ×10−31 ×v
Question 32
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron after acceleration.
Solution
Step 1: Calculate the kinetic energy of the electron using the energy gained
from the potential difference. Step 2: Apply the de Broglie wavelength formula
to find the wavelength associated with the electron.
Step 1: The kinetic energy of the electron can be calculated using the work-
energy theorem. The work done on the electron is equal to the potential energy
gained from the potential difference. This work is then converted into kinetic
energy.
The potential energy gained by the electron:
P E =qV
where qis the charge of the electron and Vis the potential difference.
Therefore, the kinetic energy of the electron:
KE =qV
Substitute q=e(charge of an electron) and V= 100 V:
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
30
Step 2: The de Broglie wavelength associated with the electron can be
calculated using the formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant, and pis the
momentum of the electron.
The momentum of the electron can be calculated using the kinetic energy:
p=√2mKE
where mis the mass of the electron and KE is the kinetic energy.
Substitute m= 9.11 ×10−31 kg and KE = 1.6×10−17 J:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈4.78 ×10−24 kg m/s
Finally, calculate the de Broglie wavelength:
λ=h
p
λ=6.63 ×10−34 J s
4.78 ×10−24 kg m/s
λ≈1.39 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration through a potential difference of 100 V is approximately 1.39 ×10−10
m.
Question 33
Question
An electron with a kinetic energy of 100 eV is moving in a straight line. Calculate
the de Broglie wavelength of the electron. (Take the mass of an electron to be
9.11 ×10−31 kg and the elementary charge to be 1.60 ×10−19 C.)
Solution
Step 1: Determine the velocity of the electron using the kinetic energy. Given
that the kinetic energy of the electron is 100 eV, we can convert this to joules
by using the conversion factor: 1 eV = 1.60 ×10−19 J. Therefore, the kinetic
energy of the electron is 100 ×1.60 ×10−19 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 vas follows:
1
2mv2= 100 ×1.60 ×10−19
31
Where mis the mass of the electron. Substitute m= 9.11 ×10−31 kg:
1
2×9.11 ×10−31 ×v2= 100 ×1.60 ×10−19
Step 3: Solve for the velocity of the electron.
v=r2×100 ×1.60 ×10−19
9.11 ×10−31
Step 4: Calculate the de Broglie wavelength using the velocity. The de
Broglie wavelength (λ) of the electron is given by:
λ=h
mv
where his the Planck constant. Substitute h= 6.63×10−34 Js, m= 9.11×10−31
kg, and the calculated value of v.
Step 5: Calculate the de Broglie wavelength.
λ=6.63 ×10−34
9.11 ×10−31 ×v
Now, substitute the value of vcalculated in Step 3 to find the de Broglie
wavelength of the electron.
Question 34
Question
An electron is accelerated through a potential difference of 150 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: We can use the de Broglie wavelength formula for matter waves, which
relates the wavelength (λ) of a particle to its momentum (p): λ=h
p, where h
is Planck’s constant (6.626 ×10−34 J·s).
Step 2: First, we need to find the momentum of the electron. The kinetic
energy of the electron can be calculated using the formula KE =eV , where e
is the elementary charge (1.6×10−19 C) and Vis the potential difference (150
V). Therefore, KE = (1.6×10−19 C)(150 V).
Step 3: The momentum of the electron can be calculated using the formula
p=√2meKE, where meis the mass of the electron (9.11 ×10−31 kg). Thus,
p=p2(9.11 ×10−31 kg)(KE).
Step 4: Substituting the expression for KE into the momentum formula, we
get p=p2(9.11 ×10−31 kg)[(1.6×10−19 C)(150 V)].
32
Step 5: Calculate the momentum pusing the above expression.
Step 6: Finally, substitute the momentum pinto the de Broglie wavelength
formula λ=h
pto find the de Broglie wavelength associated with the electron.
Question 35
Question
An electron is accelerated through a potential difference of 150 V. Determine
the de Broglie wavelength of the electron after being accelerated.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference
applied. Step 2: Use the kinetic energy to find the velocity of the electron. Step
3: Calculate the de Broglie wavelength using the velocity of the electron.
Step 1: Calculate the kinetic energy of the electron using the potential
difference applied, V= 150 V. The kinetic energy of the electron is given by
the formula:
K.E. =eV
where eis the charge of an electron and Vis the potential difference. Substitute
the values:
K.E. = (1.6×10−19 C)(150 V)
K.E. = 2.4×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy can also be expressed as:
K.E. =1
2mv2
where mis the mass of the electron and vis the velocity. Rearrange the formula
to solve for velocity:
v=r2K.E.
m
Substitute the values:
v=s2(2.4×10−17 J)
9.11 ×10−31 kg
v≈6.2×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the elec-
tron. The de Broglie wavelength is given by the formula:
λ=h
mv
33
λ=6.63 ×10−34
2.94 ×10−24 = 2.25 ×10−10 m or 0.225 nm
Therefore, the de Broglie wavelength of the electron after being accelerated
through a potential difference of 120 V is 0.225 nm.
Question 2
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with the electron.
Solution
Step 1: Determine the kinetic energy of the electron. We can use the rela-
tionship between potential energy (PE) and kinetic energy (KE) in an electron
accelerated through a potential difference:
PE = KE = eV
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(100 V). Substituting in the values:
KE = (1.6×10−19 C)(100 V) = 1.6×10−17 J
Step 2: Use the formula for kinetic energy in terms of de Broglie wavelength:
KE = 1
2mv2=h2
λ2m
where mis the mass of the electron (9.11 ×10−31 kg), vis the velocity of the
electron, λis the de Broglie wavelength, and his the Planck constant (6.63 ×
10−34 J·s).
Step 3: Rearrange the formula to solve for λ:
λ=h
√2mKE
Step 4: Substitute the values of h,m, and KE into the de Broglie wavelength
formula:
λ=6.63 ×10−34 J·s
p2(9.11 ×10−31 kg)(1.6×10−17 J)
Step 5: Calculate the de Broglie wavelength using the formula:
λ≈7.3×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 7.3×10−10 m.
2
Question 3
Question
An electron accelerated through a potential difference of 500 V has a kinetic
energy of 4.0 eV. Determine the de Broglie wavelength associated with this
electron.
Solution
Step 1: Given data: The potential difference (V) = 500 V Kinetic energy of the
electron (K.E.) = 4.0 eV
Step 2: Convert kinetic energy to joules: Since 1 eV = 1.6 x 10−19 J, the
kinetic energy in joules:
K.E. = 4.0 eV = 4.0×1.6×10−19 J=6.4×10−19 J
Step 3: Relation between kinetic energy and potential difference for an elec-
tron: The kinetic energy of an electron accelerated through a potential difference
(V) is given by:
K.E. =eV
where eis the charge of an electron (1.6 x 10−19 C).
Step 4: Solve for V:
K.E. =eV
V=K.E.
e=6.4×10−19
1.6×10−19 = 4 V
Step 5: Calculate the speed of the electron: The speed of the electron can
be calculated using the kinetic energy:
K.E. =1
2mv2
Solving for vgives:
v=r2K.E.
m
where mis the mass of the electron (9.11 x 10−31 kg).
Step 6: Substitute values and find the speed:
v=r2×6.4×10−19
9.11 ×10−31 = 8.19 ×106ms−1
Step 7: Calculate the de Broglie wavelength: The de Broglie wavelength (λ)
is given by:
λ=h
p
where his the Planck’s constant (6.626 x 10−34 Js) and pis the momentum of
the electron which is given by m×v.
3
Step 8: Substitute values and find the de Broglie wavelength:
λ=h
m×v=6.626 ×10−34
9.11 ×10−31 ×8.19 ×106= 8.02 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is 8.02 ×
10−11 m.
Question 4
Question
An electron is accelerated through a potential difference of 200 V. Calculate
the de Broglie wavelength associated with the electron after acceleration. (Take
the charge of an electron e= 1.6×10−19 C and the mass of an electron m=
9.11 ×10−31 kg.)
Solution
Step 1: Calculate the kinetic energy of the electron
Given potential difference, V= 200 V, and charge of an electron, e= 1.6×
10−19 C, the kinetic energy of the electron can be calculated by:
K.E. =eV
K.E. = (1.6×10−19 C)(200 V)
K.E. = 3.2×10−17 J
Step 2: Calculate the velocity of the electron
The kinetic energy can also be written in terms of the velocity of the electron
as:
K.E. =1
2mv2
Solving for v:
v=r2×K.E.
m
v=s2×3.2×10−17 J
9.11 ×10−31 kg
v≈6.64 ×106m/s
Step 3: Calculate the de Broglie wavelength
4
The de Broglie wavelength of the electron is given by:
λ=h
mv
where his the Planck constant (6.626 ×10−34 m2kg/s). Substituting the
values:
λ=6.626 ×10−34 m2kg/s
(9.11 ×10−31 kg)(6.64 ×106m/s)
λ≈7.29 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration is approximately 7.29 ×10−10 m.
Question 5
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =eV ,
where e= 1.6×10−19 C is the charge of an electron and V= 500 V is the
potential difference.
KE =eV
= (1.6×10−19 C)(500 V)
= 8 ×10−17 J
Step 2: Use the formula for kinetic energy to calculate the momentum of the
electron: p=√2mKE, where m= 9.11 ×10−31 kg is the mass of the electron.
p=√2mKE
=p2(9.11 ×10−31 kg)(8 ×10−17 J)
≈4.3×10−24 kg m/s
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
h= 6.63 ×10−34 J s is the Planck’s constant.
λ=h
p
=6.63 ×10−34 J s
4.3×10−24 kg m/s
≈1.54 ×10−10 m
5
Therefore, the de Broglie wavelength of the electron is approximately 1.54 ×
10−10 m.
Question 6
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 (λ) associated with the electron accelerated
through a potential difference of 100 V, we can use the de Broglie wavelength
formula:
λ=h
p=h
√2mE
where: h= Planck’s constant (6.626 ×10−34 J·s), m= mass of the electron
(9.11 ×10−31 kg), E= kinetic energy of the electron (qV where qis the charge
of the electron and Vis the potential difference).
Given that q(charge of the electron) = 1.6×10−19 C and V= 100 V, we
can calculate the kinetic energy Eand then find the de Broglie wavelength λ.
Step 1: Calculate the kinetic energy of the electron
E=qV = (1.6×10−19 C)(100 V)
E= 1.6×10−17 J
Step 2: Find the de Broglie wavelength
λ=h
√2mE =6.626 ×10−34 J·s
p2(9.11 ×10−31 kg)(1.6×10−17 J)
λ=6.626 ×10−34
√2×9.11 ×1.6×10−48
λ=6.626 ×10−34
√29.056 ×10−48
λ=6.626 ×10−34
5.39 ×10−24
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 meters.
6
Question 7
Question
An electron is accelerated through a potential difference of 150 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Find the kinetic energy of the electron using the relation KE =qV ,
where qis the charge of an electron and Vis the potential difference. Step 2:
Use the kinetic energy to find the momentum of the electron using the equation
p=√2meKE, where meis the mass of an electron. Step 3: Calculate the de
Broglie wavelength using the relation λ=h
p, where his the Planck constant.
Step 1: The charge of an electron q= 1.6×10−19 C and the potential
difference V= 150 V.
KE =qV = (1.6×10−19 C)(150 V) = 2.4×10−17 J
Step 2: The mass of an electron me= 9.11×10−31 kg.
p=p2meKE =p2(9.11 ×10−31 kg)(2.4×10−17 J) ≈4.16 ×10−24 kg m/s
Step 3: The Planck constant h= 6.63×10−34 J·s.
λ=h
p=6.63 ×10−34 J·s
4.16 ×10−24 kg m/s ≈1.59 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 1.59 ×10−10 m.
Question 8
Question
An electron is accelerated through a potential difference of 500 V. Calculate the
de Broglie wavelength of the electron after acceleration. (Mass of an electron =
9.11 ×10−31 kg, Charge of an electron = 1.60 ×10−19 C, Planck’s constant =
6.63 ×10−34 Js)
Solution
Step 1: Calculate the kinetic energy of the electron using the relation KE =qV ,
where KE is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Step 1: KE = (1.60 ×10−19 C)(500 V) = 8.00 ×10−17 J
7
Step 2: Calculate the velocity of the electron using the formula KE =1
2mv2,
where mis the mass of the electron and vis the velocity of the electron.
KE =1
2mv2
8.00 ×10−17 =1
2(9.11 ×10−31)v2
v=r2×8.00 ×10−17
9.11 ×10−31
v≈6.61 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron using the de
Broglie wavelength formula λ=h
p, where his Planck’s constant and pis the
momentum of the electron.
Step 3: p=mv = (9.11 ×10−31 kg)(6.61 ×106m/s)
p= 6.02 ×10−24 kg m/s
λ=6.63 ×10−34 J s
6.02 ×10−24 kg m/s
λ≈1.10 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration through
a potential difference of 500 V is approximately 1.10 ×10−10 m.
Question 9
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: We can use the de Broglie wavelength formula, which relates the mo-
mentum of a particle to its wavelength:
λ=h
p
where: λ= de Broglie wavelength (m), h= Planck’s constant (6.626×10−34 J·s),
p= momentum of the particle (kg m/s).
Step 2: The momentum of the electron can be calculated using its kinetic
energy. The kinetic energy of the electron can be expressed as:
K.E. =1
2mv2=eV
8
where: K.E. = kinetic energy (J), m= mass of electron (9.11 ×10−31 kg), v=
velocity of electron (m/s), e= elementary charge (1.60×10−19 C), V= potential
difference (V).
Step 3: We can rearrange the kinetic energy equation to solve for the velocity
of the electron:
v=r2eV
m
Step 4: Substituting the known values into the equation for velocity:
v=s2×1.60 ×10−19 C×100 V
9.11 ×10−31 kg
Step 5: Calculating the velocity of the electron:
v≈6.58 ×106m/s
Step 6: Now, we can calculate the momentum of the electron using the
formula p=mv.
Step 7: Substituting the mass and velocity of the electron into the momen-
tum equation:
p= (9.11 ×10−31 kg)(6.58 ×106m/s)
Step 8: Calculating the momentum of the electron:
p≈6.00 ×10−24 kg m/s
Step 9: Finally, substitute the momentum into the de Broglie wavelength
formula:
λ=(6.626 ×10−34 J·s)
6.00 ×10−24 kg m/s
Step 10: Calculating the de Broglie wavelength of the electron:
λ≈1.11 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 1.11 ×10−10 m.
Question 10
Question
An electron with a kinetic energy of 3.0×10−19 J, moving in a vacuum, exhibits
wave-like behavior. Calculate the de Broglie wavelength associated with this
electron.
9
Solution
Step 1: Recall the de Broglie wavelength formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34
Js), and pis the momentum of the particle.
Step 2: We can find the momentum of the electron using the formula:
K.E. =1
2mv2=p2
2m
where mis the mass of the electron and vis its velocity.
Step 3: Since v=q2K.E.
m, we can substitute this into the momentum
formula to get:
p=√2mK.E.
Step 4: Now we substitute pinto the de Broglie wavelength formula:
λ=h
√2mK.E.
Step 5: Plugging in the known values, we have:
λ=6.626 ×10−34 Js
p2×9.11 ×10−31 kg ×3.0×10−19 J
Step 6: Solving this expression gives:
λ=6.626 ×10−34 Js
p2×9.11 ×10−31 kg ×3.0×10−19 J= 2.74 ×10−12 m
Therefore, the de Broglie wavelength associated with the electron is 2.74 ×
10−12 m.
Question 11
Question
An electron is accelerated through a potential difference of 50 V. Determine the
de Broglie wavelength associated with the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential differ-
ence. Step 2: Utilize the de Broglie wavelength formula to find the wavelength
associated with the electron.
10
Step 1: Calculate the kinetic energy of the electron using the potential
difference.
The kinetic energy of an electron accelerated through a potential difference
can be calculated using the equation:
K.E. =eV
where eis the elementary charge (1.6×10−19 C) and Vis the potential difference
(50 V).
Substitute the values to find the kinetic energy:
K.E. = (1.6×10−19 C) ×(50 V) = 8 ×10−18 J
Step 2: Utilize the de Broglie wavelength formula to find the wavelength
associated with the electron.
The de Broglie wavelength (λ) of a particle with kinetic energy K.E. and
mass mcan be calculated using the formula:
λ=h
p=h
√2mK.E.
where his the Planck constant (6.626 ×10−34 m2kg/s) and mis the mass of the
electron (9.109 ×10−31 kg).
Substitute the values into the formula to find the de Broglie wavelength:
λ=6.626 ×10−34 m2kg/s
p2×9.109 ×10−31 kg ×8×10−18 J
λ=6.626 ×10−34 m2kg/s
√1.451 ×10−11
λ≈6.626 ×10−34 m2kg/s
1.205 ×10−5
λ≈5.496 ×10−30 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 50 V is approximately 5.496 ×10−30 m.
Question 12
Question
An electron is accelerated through a potential difference of 1000 V. Determine
the de Broglie wavelength of the electron.
11
Solution
Step 1: The kinetic energy of the electron can be calculated using the formula:
KE =eV
where eis the charge of an electron (1.6 x 10−19 C), and Vis the potential
difference (1000 V in this case). Therefore,
KE = (1.6×10−19 C)(1000 V)
KE = 1.6×10−16 J
Step 2: The velocity of the electron can be determined using the kinetic
energy formula:
KE =1
2mv2
where mis the mass of an electron (9.11 x 10−31 kg). Therefore,
1.6×10−16 =1
2(9.11 ×10−31)v2
v2=2(1.6×10−16)
9.11 ×10−31
v≈3.53 ×106m/s
Step 3: The de Broglie wavelength of the electron can be calculated using
the formula:
λ=h
p=h
mv
where his the Planck constant (6.63 ×10−34 J·s). Substituting the values,
λ=6.63 ×10−34
(9.11 ×10−31)(3.53 ×106)
λ≈2.47 ×10−10 m
Therefore, the de Broglie wavelength of the electron accelerated through a
potential difference of 1000 V is approximately 2.47 ×10−10 m.
Question 13
Question
An electron is accelerated through a potential difference of 200 V. Determine
the de Broglie wavelength associated with the electron’s motion.
12
Solution
Step 1: Recall 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 particle.
Step 2: Next, we need to find the momentum of the electron. The kinetic
energy gained by the electron is given by:
K.E. =qV
where qis the charge of the electron (1.6×10−19 C) and Vis the potential
difference (200 V).
Step 3: The kinetic energy can also be expressed in terms of momentum as:
K.E. =p2
2m
where mis the mass of the electron (9.11 ×10−31 kg).
Step 4: Equating the two expressions for kinetic energy, we have:
qV =p2
2m
Step 5: Solving for p, we get:
p=p2m·qV
Step 6: Substitute the known values to find p:
p=p2·9.11 ×10−31 kg ·1.6×10−19 C·200 V
Step 7: Calculate pto get:
p≈1.48 ×10−24 kg ·m/s
Step 8: Finally, substitute pinto the de Broglie wavelength formula to find
λ:
λ=6.63 ×10−34 J·s
1.48 ×10−24 kg ·m/s
Step 9: Calculate λto get the de Broglie wavelength associated with the
electron’s motion.
Question 14
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength associated with this electron.
13
Solution
Step 1: Start with the formula for de Broglie wavelength:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant (6.626 ×10−34 J
s), and pis the momentum of the particle.
Step 2: Calculate the momentum of the electron from its kinetic energy
gained by passing through the potential difference:
K.E. =e×V=1
2mv2
where eis the charge of an electron (1.6×10−19 C), Vis the potential difference
(100 V), mis the mass of an electron (9.11 ×10−31 kg), and vis the velocity of
the electron.
Step 3: The velocity of an electron with kinetic energy K.E. can be expressed
as:
v=r2×K.E.
m
Step 4: Substitute the given values and solve for v:
v=r2×(e×V)
m
Step 5: Once you have the velocity, you can calculate the momentum pusing
p=mv.
Step 6: Finally, substitute the momentum pinto the de Broglie wavelength
formula to find the wavelength λ. Remember to convert the momentum to SI
units (kg m/s) if necessary.
Question 15
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 electron, e= 1.6×10−19 C Mass of electron, m= 9.11 ×
10−31 kg Planck’s constant, h= 6.63 ×10−34 J·s Speed of light, c= 3.00 ×108
m/s
Solution
Step 1: Calculate the kinetic energy of the electron after acceleration. Given
that the potential difference is 100 V, the kinetic energy can be calculated using
14
the formula: KE =qV , where qis the charge and Vis the potential difference.
Calculating the value of KE:
KE = 1.6×10−19 C×100 V = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using the kinetic energy. The
kinetic energy can be expressed as:
KE =1
2mv2
where mis the mass of the electron and vis the velocity. Solving for v:
v=r2KE
m=s2×1.6×10−17 J
9.11 ×10−31 kg ≈3.93 ×106m/s
Step 3: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength is given by the formula:
λ=h
mv
Substitute the values of h,m, and vinto the formula:
λ=6.63 ×10−34 J·s
9.11 ×10−31 kg ×3.93 ×106m/s
λ≈1.70 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.70 ×10−10 m.
Question 16
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
given. Step 2: Use the kinetic energy to find the momentum of the electron.
Step 3: Apply de Broglie’s wavelength formula to determine the de Broglie
wavelength.
Step 1: The kinetic energy of the electron can be calculated using the
formula:
Kinetic energy (KE) = q·potential difference
15
where qis the charge of the electron. The charge of an electron is −1.6×10−19
C. Substituting the values:
KE = (−1.6×10−19 C) ×(100 V)
KE = −1.6×10−17 J
Step 2: The kinetic energy of the electron can also be expressed in terms
of momentum pas:
KE = p2
2m
where mis the mass of the electron. The mass of an electron is 9.11 ×10−31
kg. Rearranging the equation to solve for momentum:
p=√2m·KE = p2×9.11 ×10−31 kg ×1.6×10−17 J
p=√2×9.11 ×1.6×10−17
p≈8.55 ×10−17 kg m/s
Step 3: De Broglie’s wavelength formula is given by:
λ=h
p
where his the Planck constant (6.63 ×10−34 J s). Substituting the values:
λ=6.63 ×10−34 J s
8.55 ×10−17 kg m/s
λ≈7.77 ×10−11 m
Therefore, the de Broglie wavelength of the electron is approximately 7.77 ×
10−11 m.
Question 17
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron after acceleration. (Mass of an electron =
9.11 ×10−31 kg, charge of an electron = −1.6×10−19 C)
16
Solution
Step 1: First, we calculate the kinetic energy of the electron using the formula
KE =qV , where qis the charge and Vis the potential difference.
Step 1: KE =qV = (−1.6×10−19 C)(100 V) = −1.6×10−17 J
Step 2: Next, we calculate the velocity of the electron using the formula for
kinetic energy: KE =1
2mv2, where mis the mass of the electron.
Step 2: KE =1
2mv2=⇒v=r2·KE
m=s2·1.6×10−17 J
9.11 ×10−31 kg ≈2.44×106m/s
Step 3: Now, we can calculate the de Broglie wavelength of the electron using
the formula: λ=h
p, where his the Planck constant and pis the momentum.
Step 3: p=mv = (9.11 ×10−31 kg)(2.44 ×106m/s) ≈2.22 ×10−24 kg m/s
λ=h
p=6.63 ×10−34 J s
2.22 ×10−24 kg m/s ≈2.99 ×10−9m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 2.99 ×10−9m.
Question 18
Question
A beam of electrons is accelerated through a potential difference of 200 V.
Calculate the de Broglie wavelength of these electrons.
Solution
Step 1: Determine the kinetic energy of the electrons using the potential differ-
ence.
The kinetic energy of an electron accelerated through a potential difference
is given by the formula:
Ekin =qV
where qis the charge of an electron (−1.6×10−19 C) and Vis the potential
difference (200 V).
Ekin = (−1.6×10−19 C)(200 V)
Ekin =−3.2×10−17 J
Step 2: Calculate the momentum of the electrons.
17
The momentum of an electron can be calculated using the formula:
p=p2mEkin
where mis the mass of an electron (9.11 ×10−31 kg) and Ekin is the kinetic
energy calculated in Step 1.
p=p2(9.11 ×10−31 kg)(−3.2×10−17 J)
p≈1.32 ×10−24 kg m/s
Step 3: Find the de Broglie wavelength of the electrons using the formula:
λ=h
p
where his the Planck constant (6.626 ×10−34 J s) and pis the momentum
calculated in Step 2.
λ=6.626 ×10−34 J s
1.32 ×10−24 kg m/s
λ≈5.02 ×10−10 m
Therefore, the de Broglie wavelength of the electrons accelerated through a
potential difference of 200 V is approximately 5.02 ×10−10 m.
Question 19
Question
An electron with a kinetic energy of 200 eV is moving through space. Determine
the de Broglie wavelength associated with the electron.
Solution
Step 1: Convert the kinetic energy of the electron from electronvolts (eV) to
joules (J). Step 2: Use the de Broglie wavelength formula to find the wavelength
associated with the electron.
Step 1: Given: Kinetic energy of the electron = 200 eV
We know that 1 electronvolt is equal to 1.6×10−19 J. Therefore, the kinetic
energy of the electron in joules is:
200 eV ×1.6×10−19 J/eV = 3.2×10−17 J
Step 2: The de Broglie wavelength of a particle is given by:
λ=h
p
18
where λis the de Broglie wavelength, his the Planck constant (6.626×10−34 J s),
and pis the momentum of the particle.
The momentum of the electron can be calculated using the equation:
p=√2mK
where mis the mass of the electron (9.11×10−31 kg) and Kis the kinetic energy.
Substitute the values into the equation to find the momentum of the electron:
p=p2×9.11 ×10−31 kg ×3.2×10−17 J≈7.57 ×10−24 kg m/s
Now substitute the momentum into the de Broglie wavelength formula to
find the wavelength:
λ=6.626 ×10−34 J s
7.57 ×10−24 kg m/s ≈8.75 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is 8.75 ×
10−11 meters.
Question 20
Question
An electron is accelerated through a potential difference of 100 volts. Calculate
the de Broglie wavelength of the electron.
Solution
Step 1: We can start by finding the kinetic energy of the electron using the
formula K=eV , where eis the elementary charge (1.6×10−19 coulombs) and
Vis the potential difference (100 volts). Step 2: Substitute the values into the
formula to find the kinetic energy K. Step 3: Next, we can use the de Broglie
wavelength formula λ=h
p, where his the Planck constant (6.63×10−34 J·s) and
pis the momentum of the electron. Step 4: The momentum pcan be calculated
using the equation p=√2mK, where mis the mass of the electron (9.11×10−31
kg) and Kis the kinetic energy. Step 5: Substitute the kinetic energy Kinto
the momentum equation to find the momentum p. Step 6: Finally, substitute
the momentum pinto the de Broglie wavelength formula to calculate the de
Broglie wavelength λ.
Question 21
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron.
19
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 mis
the mass of the electron. Step 3: Use the de Broglie wavelength formula λ=h
p,
where his the Planck constant and pis the momentum of the electron. Step
4: Calculate the momentum of the electron using the formula p=mv. Step
5: Substitute the momentum of the electron into the de Broglie wavelength
formula to find the de Broglie wavelength.
Step 1: Calculate the kinetic energy of the electron. The elementary charge
e= 1.6×10−19 C. The potential difference V= 100 V. The kinetic energy
K=eV = (1.6×10−19 C)(100 V) = 1.6×10−17 J.
Step 2: Calculate the velocity of the electron. The mass of the electron
m= 9.11 ×10−31 kg. Using the kinetic energy formula: K=1
2mv21.6×
10−17 =1
2(9.11 ×10−31)(v2)v2=2(1.6×10−17 )
9.11×10−31 v2≈3.53 ×106v≈√3.53 ×106
v≈1.88 ×103m/s.
Step 3: Use the de Broglie wavelength formula. The Planck constant
h= 6.63 ×10−34 J s. The momentum of the electron p=mv = (9.11 ×
10−31 kg)(1.88 ×103m/s) = 1.71 ×10−27 kg m/s.
Step 4: Calculate the de Broglie wavelength. λ=h
p=6.63×10−34 J s
1.71×10−27 kg m/s
λ≈3.88 ×10−10 m.
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 100 V is approximately 3.88 ×10−10 m.
Question 22
Question
An electron is accelerated through a potential difference of 120 V. Determine
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 ×10−19 C) and Vis the potential
difference (120 V).
KE = (1.6×10−19 C) ×(120 V)
Step 2: Calculating the kinetic energy:
KE = 1.92 ×10−17 J
Step 3: Next, we can relate the kinetic energy of the electron to its de Broglie
wavelength using the formula:
KE =1
2mv2=h2
λ2·2m
20
Step 4: Rearranging the equation to solve for λ:
λ=h
√2mKE
Step 5: Plugging in the values of Planck’s constant h(6.626 ×10−34 J·s)
and the mass of an electron m(9.11 ×10−31 kg), and the calculated kinetic
energy:
λ=6.626 ×10−34 J·s
p2×9.11 ×10−31 kg ×1.92 ×10−17 J
Step 6: Solving for λ:
λ≈2.3×10−12 m
Therefore, the de Broglie wavelength associated with the electron accelerated
through a potential difference of 120 V is approximately 2.3×10−12 m.
Question 23
Question
An electron is accelerated through a potential difference of 100 V. Calculate
the de Broglie wavelength of the electron after acceleration. (Take the electron
charge as 1.6×10−19 C and its mass as 9.11 ×10−31 kg)
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.
Given: q= 1.6×10−19 C, V = 100 V
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
Step 2: Calculate the velocity of the electron using the formula for kinetic
energy: KE =1
2mv2, where mis the mass of the electron.
Given: m= 9.11 ×10−31 kg
1.6×10−17 =1
2(9.11 ×10−31)v2
v2=2×1.6×10−17
9.11 ×10−31
v2=3.2×10−17
9.11 ×10−31
21
v2≈3.51 ×1013
v≈p3.51 ×1013
v≈5.92 ×106m/s
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
his the Planck’s constant and pis the momentum of the electron.
Given: h= 6.63 ×10−34 J s
p=m·v= (9.11 ×10−31 kg)(5.92 ×106m/s)
p≈5.40 ×10−24 kg m/s
λ=6.63 ×10−34 J s
5.40 ×10−24 kg m/s
λ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 1.23 ×10−10 m.
Question 24
Question
An electron with a mass of 9.11 ×10−31 kg is accelerated through a potential
difference of 80 V. Calculate the de Broglie wavelength of the electron.
Solution
Step 1: Calculate the kinetic energy of the electron using the formula KE =qV ,
where qis the charge of the electron (we’ll assume it is −1.6×10−19 C) and V
is the potential difference.
KE = (1.6×10−19 C)(80 V)
KE = 1.28 ×10−17 J
Step 2: Use the kinetic energy to find the speed of the electron using the
formula KE =1
2mv2, where mis the mass of the electron.
1.28 ×10−17 =1
2×(9.11 ×10−31)v2
v2=2×1.28 ×10−17
9.11 ×10−31
v≈3.04 ×106m/s
22
Step 3: Calculate the de Broglie wavelength using the formula λ=h
p, where
his Planck’s constant and pis the momentum. The momentum can be calcu-
lated as mv.
p= (9.11 ×10−31)(3.04 ×106)
p≈2.77 ×10−24 kg m/s
λ=6.626 ×10−34
2.77 ×10−24
λ≈2.39 ×10−10 m
Therefore, the de Broglie wavelength of the electron is 2.39 ×10−10 m.
Question 25
Question
An electron is accelerated through a potential difference of 250 V. Calculate the
de Broglie wavelength associated with this electron.
Solution
Step 1: Use the energy gained by the electron to find its kinetic energy. Step 2:
Apply the de Broglie wavelength formula to calculate the wavelength.
Step 1: Calculate the kinetic energy of the electron. The energy gained
by the electron is equal to the potential energy difference it was accelerated
through, which can be converted to kinetic energy. The kinetic energy is given
by the formula:
K=qV
where qis the charge of the electron (1.6 x 10-19 C) and Vis the potential
difference (250 V).
K= (1.6×10−19 C) ×(250 V)
K= 4 ×10−17 J
Step 2: Calculate the de Broglie wavelength. The de Broglie wavelength is
given by:
λ=h
p
where: h= Planck’s constant = 6.626 x 10-34 J
·
s, p= momentum of the electron,
and λ= de Broglie wavelength.
The momentum of the electron can be calculated using the kinetic energy:
p=√2mK
where m= mass of the electron = 9.11 x 10-31 kg and K= kinetic energy.
23
Substitute the values to find p:
p=p2×9.11 ×10−31 kg ×4×10−17 J
p≈1.19 ×10−23 kg m/s
Now, substitute hand pinto the de Broglie wavelength formula:
λ=6.626 ×10−34 J
·
s
1.19 ×10−23 kg m/s
λ≈5.57 ×10−11 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 5.57 ×10−11 meters.
Question 26
Question
A particle of mass mmoves with velocity vand has kinetic energy E. Determine
the de Broglie wavelength of this particle.
Solution
The de Broglie wavelength of a particle is given by the formula:
λ=h
p
where his the Planck constant (6.62607015 ×10−34 Js) and pis the momentum
of the particle.
Step 1: First, we need to express the momentum of the particle in terms of
its kinetic energy and mass. The momentum of the particle is given by:
p=mv
Step 2: Next, we can express the velocity vin terms of kinetic energy E
and mass m. The kinetic energy of the particle is given by:
E=1
2mv2
Solving for vgives:
v=r2E
m
Step 3: Substituting vback into the momentum formula, we get:
p=mr2E
m=√2mE
24
Step 4: Now, we can substitute the momentum pinto the formula for de
Broglie wavelength:
λ=h
√2mE =h
q2m·1
2mv2
=h
√mE
Hence, the de Broglie wavelength of the particle is h
√mE .
Question 27
Question
An electron is accelerated from rest through a potential difference of 1000 V.
Calculate the de Broglie wavelength of the electron after acceleration.
Solution
Step 1: Find the kinetic energy of the electron using the potential difference.
The kinetic energy of the electron can be calculated using the formula:
K.E. =e·V
where eis the elementary charge and Vis the potential difference. Given that
the potential difference is 1000 V, and the elementary charge e= 1.6×10−19
C, we have:
K.E. = 1.6×10−19 ×1000
K.E. = 1.6×10−16 J
Step 2: Use the kinetic energy to find the de Broglie wavelength. The de
Broglie wavelength of a particle is given by the formula:
λ=h
p
where his the Planck constant (6.626 ×10−34 m2kg/s) and pis the momentum
of the particle. The momentum of the electron can be calculated using the
kinetic energy:
p=√2mK.E.
where mis the mass of the electron (9.11 ×10−31 kg) and K.E. is the kinetic
energy. Substitute the known values into the momentum formula:
p=p2×9.11 ×10−31 ×1.6×10−16
p≈2.63 ×10−24 kg m/s
Now, substitute the momentum into the de Broglie wavelength formula:
λ=6.626 ×10−34
2.63 ×10−24
25
λ≈2.51 ×10−12 m
Therefore, the de Broglie wavelength of the electron after acceleration is
approximately 2.51 ×10−12 meters.
Question 28
Question
A neutron is accelerated through a potential difference of 50 V. Calculate the
de Broglie wavelength associated with the neutron after being accelerated. The
mass of a neutron is 1.67 ×10−27 kg and the charge of an electron is 1.6×10−19
C.
Solution
Step 1: Calculate the kinetic energy gained by the neutron. Given that the
potential difference is 50 V, the kinetic energy gained by the neutron can be
calculated using the equation:
K=qV
where: K= kinetic energy gained (J), q= charge of the particle (C), V=
potential difference (V).
Substitute the values:
K= (1.6×10−19C)×(50V)=8×10−18J
Step 2: Calculate the velocity of the neutron. The kinetic energy gained is
equal to the energy associated with the motion of the neutron, so we have:
K=1
2mv2
Solve for velocity (v):
v=r2K
m
v=s2×8×10−18J
1.67 ×10−27kg
v=r1.6×10−17
1.67 ×1010
v≈2.39 ×105m/s
Step 3: Calculate the de Broglie wavelength associated with the neutron.
The de Broglie wavelength (λ) is given by:
λ=h
p
26
where: h= Planck’s constant (6.63×10−34 J·s), p= momentum of the neutron
(m×v).
Substitute the values for momentum:
p= 1.67 ×10−27 kg ×2.39 ×105m/s
p= 4.0×10−22 kg m/s
Now, calculate the de Broglie wavelength:
λ=6.63 ×10−34 J·s
4.0×10−22 kg m/s
λ=6.63
4.0×10−12 m
λ≈1.66 ×10−12 m
Therefore, the de Broglie wavelength associated with the neutron after being
accelerated is approximately 1.66 ×10−12 m.
Question 29
Question
Consider an electron with a kinetic energy of 100 eV. If the electron behaves as
a wave, what is 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 J
s), and pis the momentum of the electron.
First, we need to find the momentum of the electron using the kinetic energy.
Given that the kinetic energy of the electron is 100 eV, we need to convert this
energy to joules:
1 eV = 1.602 ×10−19 joules
Thus, the kinetic energy in joules is:
100 eV ×1.602 ×10−19 J/eV = 1.602 ×10−17 J
Next, we can use the formula for kinetic energy to find the momentum p:
KE =p2
2m
27
where mis the mass of an electron (9.11 ×10−31 kg).
Step 1: Solve for momentum pin the kinetic energy formula:
p=√2mKE
p=p2×9.11 ×10−31 ×1.602 ×10−17
p=p2.9×10−32
p≈5.39 ×10−16 kg m/s
Now that we have the momentum of the electron, we can find the de Broglie
wavelength:
Step 2: Calculate the de Broglie wavelength:
λ=h
p
λ=6.626 ×10−34
5.39 ×10−16
λ≈1.23 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron is approx-
imately 1.23 ×10−10 meters.
Question 30
Question
A particle of mass mis confined to move in one dimension within a box of
length L. The energy of the particle is given by E=n2ℏ2π2
2mL2, where nis a
positive integer. Determine the de Broglie wavelength of the particle in terms
of Land n.
Solution
Step 1: The de Broglie wavelength of a particle is given by λ=h
p, where his
the Planck constant and pis the momentum of the particle.
Step 2: The momentum of the particle can be expressed in terms of its
energy Eusing the relation E=p2
2m. Rearranging, we have p=√2mE.
Step 3: Substitute the given expression for energy into the momentum equa-
tion to find the momentum in terms of n,ℏ,m, and L.
p=r2m·n2ℏ2π2
2mL2=nℏπ
L
28
Step 4: Now, substitute the expression for momentum into the de Broglie
wavelength formula to find the de Broglie wavelength λ.
λ=h
p=h
nℏπ/L =hL
nℏπ
Step 5: Simplify the expression to obtain the final result for the de Broglie
wavelength in terms of Land n.
λ=L
nπ
Question 31
Question
An electron is accelerated through a potential difference of 100 V. Calculate the
de Broglie wavelength of the electron in meters. Given that the charge of an
electron is −1.6×10−19 C and Planck’s constant is 6.63 ×10−34 J·s.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference.
The kinetic energy of the electron can be calculated using the formula:
K.E. =qV
where K.E. is the kinetic energy, qis the charge of the electron, and Vis the
potential difference.
Substitute q=−1.6×10−19 C and V= 100 V into the formula:
K.E. =−1.6×10−19 C×100 V
Step 2: Calculate the velocity of the electron. The kinetic energy of an
electron can be expressed in terms of its velocity using the formula:
K.E. =1
2mv2
where K.E. is the kinetic energy, mis the mass of the electron, and vis the
velocity.
Equating the two expressions for kinetic energy:
−1.6×10−19 ×100 = 1
2×9.11 ×10−31 ×v2
Step 3: Solve for the velocity of the electron.
v=r−1.6×10−19 ×100 ×2
9.11 ×10−31
29
Step 4: Calculate the de Broglie wavelength of the electron. The de Broglie
wavelength of an electron can be calculated using the formula:
λ=h
p
where λis the de Broglie wavelength, his Planck’s constant, and pis the
momentum of the electron.
The momentum of the electron can be expressed as:
p=mv
Substitute m= 9.11 ×10−31 kg, vcalculated in step 3, and h= 6.63 ×10−34
J·s into the formula:
λ=6.63 ×10−34
9.11 ×10−31 ×v
Question 32
Question
An electron is accelerated through a potential difference of 100 V. Determine
the de Broglie wavelength associated with the electron after acceleration.
Solution
Step 1: Calculate the kinetic energy of the electron using the energy gained
from the potential difference. Step 2: Apply the de Broglie wavelength formula
to find the wavelength associated with the electron.
Step 1: The kinetic energy of the electron can be calculated using the work-
energy theorem. The work done on the electron is equal to the potential energy
gained from the potential difference. This work is then converted into kinetic
energy.
The potential energy gained by the electron:
P E =qV
where qis the charge of the electron and Vis the potential difference.
Therefore, the kinetic energy of the electron:
KE =qV
Substitute q=e(charge of an electron) and V= 100 V:
KE = (1.6×10−19 C)(100 V)
KE = 1.6×10−17 J
30
Step 2: The de Broglie wavelength associated with the electron can be
calculated using the formula:
λ=h
p
where λis the de Broglie wavelength, his the Planck constant, and pis the
momentum of the electron.
The momentum of the electron can be calculated using the kinetic energy:
p=√2mKE
where mis the mass of the electron and KE is the kinetic energy.
Substitute m= 9.11 ×10−31 kg and KE = 1.6×10−17 J:
p=p2(9.11 ×10−31 kg)(1.6×10−17 J)
p≈4.78 ×10−24 kg m/s
Finally, calculate the de Broglie wavelength:
λ=h
p
λ=6.63 ×10−34 J s
4.78 ×10−24 kg m/s
λ≈1.39 ×10−10 m
Therefore, the de Broglie wavelength associated with the electron after ac-
celeration through a potential difference of 100 V is approximately 1.39 ×10−10
m.
Question 33
Question
An electron with a kinetic energy of 100 eV is moving in a straight line. Calculate
the de Broglie wavelength of the electron. (Take the mass of an electron to be
9.11 ×10−31 kg and the elementary charge to be 1.60 ×10−19 C.)
Solution
Step 1: Determine the velocity of the electron using the kinetic energy. Given
that the kinetic energy of the electron is 100 eV, we can convert this to joules
by using the conversion factor: 1 eV = 1.60 ×10−19 J. Therefore, the kinetic
energy of the electron is 100 ×1.60 ×10−19 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 vas follows:
1
2mv2= 100 ×1.60 ×10−19
31
Where mis the mass of the electron. Substitute m= 9.11 ×10−31 kg:
1
2×9.11 ×10−31 ×v2= 100 ×1.60 ×10−19
Step 3: Solve for the velocity of the electron.
v=r2×100 ×1.60 ×10−19
9.11 ×10−31
Step 4: Calculate the de Broglie wavelength using the velocity. The de
Broglie wavelength (λ) of the electron is given by:
λ=h
mv
where his the Planck constant. Substitute h= 6.63×10−34 Js, m= 9.11×10−31
kg, and the calculated value of v.
Step 5: Calculate the de Broglie wavelength.
λ=6.63 ×10−34
9.11 ×10−31 ×v
Now, substitute the value of vcalculated in Step 3 to find the de Broglie
wavelength of the electron.
Question 34
Question
An electron is accelerated through a potential difference of 150 V. Determine
the de Broglie wavelength associated with this electron.
Solution
Step 1: We can use the de Broglie wavelength formula for matter waves, which
relates the wavelength (λ) of a particle to its momentum (p): λ=h
p, where h
is Planck’s constant (6.626 ×10−34 J·s).
Step 2: First, we need to find the momentum of the electron. The kinetic
energy of the electron can be calculated using the formula KE =eV , where e
is the elementary charge (1.6×10−19 C) and Vis the potential difference (150
V). Therefore, KE = (1.6×10−19 C)(150 V).
Step 3: The momentum of the electron can be calculated using the formula
p=√2meKE, where meis the mass of the electron (9.11 ×10−31 kg). Thus,
p=p2(9.11 ×10−31 kg)(KE).
Step 4: Substituting the expression for KE into the momentum formula, we
get p=p2(9.11 ×10−31 kg)[(1.6×10−19 C)(150 V)].
32
Step 5: Calculate the momentum pusing the above expression.
Step 6: Finally, substitute the momentum pinto the de Broglie wavelength
formula λ=h
pto find the de Broglie wavelength associated with the electron.
Question 35
Question
An electron is accelerated through a potential difference of 150 V. Determine
the de Broglie wavelength of the electron after being accelerated.
Solution
Step 1: Calculate the kinetic energy of the electron using the potential difference
applied. Step 2: Use the kinetic energy to find the velocity of the electron. Step
3: Calculate the de Broglie wavelength using the velocity of the electron.
Step 1: Calculate the kinetic energy of the electron using the potential
difference applied, V= 150 V. The kinetic energy of the electron is given by
the formula:
K.E. =eV
where eis the charge of an electron and Vis the potential difference. Substitute
the values:
K.E. = (1.6×10−19 C)(150 V)
K.E. = 2.4×10−17 J
Step 2: Use the kinetic energy to find the velocity of the electron. The
kinetic energy can also be expressed as:
K.E. =1
2mv2
where mis the mass of the electron and vis the velocity. Rearrange the formula
to solve for velocity:
v=r2K.E.
m
Substitute the values:
v=s2(2.4×10−17 J)
9.11 ×10−31 kg
v≈6.2×106m/s
Step 3: Calculate the de Broglie wavelength using the velocity of the elec-
tron. The de Broglie wavelength is given by the formula:
λ=h
mv
33
where his Planck’s constant, mis the mass of the electron, and vis the velocity.
Substitute the values:
λ=6.626 ×10−34 J s
(9.11 ×10−31 kg)(6.2×106m/s)
λ≈1.11 ×10−10 m
Therefore, the de Broglie wavelength of the electron after being accelerated
through a potential difference of 150 V is approximately 1.11 ×10−10 m.
34
