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Class- XII-CBSE-Physics Dual Nature of Radiation and Matter
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CBSE NCERT Solutions for Class 12 Physics Chapter 11 Back of Chapter Questions
1. Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given,
The potential of the electron isππ = 30kV = 3 Γ 104ππ.
The potential energy in electron volts is πΈπΈ = 3 Γ 104eV.
(a) Let the maximum frequency of X-rays produced be ππ.
The equation for frequency is,
πΈπΈ = βππ
β ππ =πΈπΈβ
Substituting the values,
ππ =1.6 Γ 10β19 Γ 3 Γ 104
6.626 Γ 10β34= 7.24 Γ 1018Hz
Thus, the maximum frequency of X-rays produced is 7.24 Γ 1018 Hz.
(b) The equation for minimum wavelength of X-rays produced is
ππ =ππππ
Substituting the values,
ππ =3 Γ 108
7.24 Γ 1018
= 4.14 Γ 10β11 m
= 0.0414 nm
Thus, the minimum wavelength of X-rays produced is 0.0414 nm.
2. The work function of caesium metal is 2.14 eV. When light of frequency 6 Γ 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
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(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Solution:
Given
Here, the work function of caesium metal is 2.14 eV.
The frequency of light is 6 Γ1014 Hz.
(a) For the emitted electrons, the maximum kinetic energy is given by the equation,
πΎπΎ = βππ β ππ0
Substituting the values,
πΎπΎ =(6.626 Γ 10β34)(6 Γ 1014)
1.6 Γ 10β19β 2.14
= 0.345 eV
Thus, the maximum kinetic energy of the emitted electrons is 0.345 eV.
(b) The stopping potential of the emitted electron is found from the relation,
πΎπΎ = ππππ0
β ππ0 =πΎπΎππ
Substituting the values,
ππ0 =0.345 Γ 1.6 Γ 10β19
1.6 Γ 10β19 = 0.345 V
Thus, the stopping potential of the emitted electron is 0.345 V.
(c) The maximum speed of the photoelectrons is found from the relation,
π£π£2 =2πΎπΎππ
β π£π£ = οΏ½2πΎπΎππ
Substituting the values,
π£π£ = οΏ½2 Γ 0.345 Γ 1.6 Γ 10β19
9.1 Γ 10β31
= β0.1104 Γ 1012
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= 3.323 Γ 105 m/s
= 332.3 km/s
Thus, the maximum speed of the emitted photoelectrons is 332.3 km/s.
3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Solution:
Given
The cutoff voltage is ππ0 = 1.5 V.
The maximum kinetic energy of the electrons is given by the equation,
πΎπΎ = ππππ0
Substituting the values,
πΎπΎ = 1.6 Γ 10β19 Γ 1.5 = 2.4 Γ 10β19 J
Thus, the maximum kinetic energy of the emitted electrons is 2.4 Γ 10β19 J.
4. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel to have the same momentum as that of the photon?
Solution:
Given
The wavelength of the monochromatic light is ππ = 632.8 nm = 632.8 Γ 10β9 m.
The emitted power of the laser is ππ = 9.42 mW = 9.42 Γ 10β3 W.
The mass of the hydrogen atom is, ππ = 1.66 Γ 10β27 kg.
(a) The energy of each photon in the light beam is
πΈπΈ =βππππ
=6.626 Γ 10β34 Γ 3 Γ 108
632.8 Γ 10β9
= 3.141 Γ 10β19 J
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The momentum of each photon in the light beam is
ππ =βππ
=6.626 Γ 10β34
632.8 Γ 10β9
= 1.047 Γ 10β27 kg m/s
Thus, the energy of each photon in the light beam is 3.141 Γ 10β19 J and the momentum of each photon in the light beam is 1.047 Γ 10β27 kg m/s.
(b) Let the number of photons arriving per second at the target be n.
The number can be found from the equation for power.
ππ = πππΈπΈ
β ππ =πππΈπΈ
=9.42 Γ 10β3
3.141 Γ 10β19= 3 Γ 1016 photons/s
Thus, the number of photons arriving per second at the target is 3 Γ 1016 photons/s.
(c) Momentum of photon and hydrogen atom is the same.
ππ = 1.047 Γ 10β27 kgm/s
The speed of the hydrogen atom is,
π£π£ =ππππ
=1.047 Γ 10β27
1.66 Γ 10β27= 0.621 m/s
Thus, the speed of the hydrogen atom is 0.621 m/s.
5. The energy flux of sunlight reaching the surface of the earth is 1.388 Γ 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given
The energy flux of sunlight is Ξ¦ = 1.388 Γ 103 W/m2
Average wavelength of the photon is ππ = 550 nm = 550 Γ 10β9 m.
The power of sunlight per square metre is ππ = 1.388 Γ 103 W
Let the number of photons incident per second on earth be ππ.
ππ = πππΈπΈ
β ππ =πππΈπΈ
=ππππβππ
Substituting the values,
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ππ =(1.388 Γ 103)(550 Γ 10β9)(6.626 Γ 10β34)(3 Γ 108)
β ππ = 3.84 Γ 1021 photons/m2/s
Thus, the number of photons (nearly) per square metre are incident on the earth per second is 3.84 Γ 1021 photons.
6. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 Γ 10β15 V s. Calculate the value of Planckβs constant.
Solution:
The slope of cut-off voltage versus frequency of incident light is,
ππππ = 4.12 Γ 10β15 V s
The relationship between the cut-off voltage and frequency is given by,
βππ = ππππ
Here,
ππ is charge of electron
β is the Plankβs constant
Therefore, the Plankβs constant is,
β = ππππππ
= 1.6 Γ 10β19 C Γ 4.12 Γ 10β15
= 6.59 Γ 10β34 J s
Therefore, the value of Plankβs constant is 6.59 Γ 10β34 J s.
7. A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Solution:
The power of the sodium lamp is, ππ = 100 W
The wavelength of sodium light emitted is,
ππ = 589 nm = 589 Γ 10β9 m
The Plankβs constant, β = 6.626 Γ 10β34 J s
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The speed of light, ππ = 3 Γ 108 m/s
(a) The energy per photon for the sodium light is,
πΈπΈ =βππππ
=6.626 Γ 10β34 Js Γ 3 Γ 10β8 m/s
589 Γ 10β9 m
= 3.37 Γ 10β19 J
= (3.37 Γ 10β19 J) Γ1 eV
1.6 Γ 10β19 J
= 2.11 eV
(b) The power is,
ππ = πππΈπΈ
Here, ππ is the number of photons delivered to the sphere.
Therefore, the number of photons delivered to the sphere is,
ππ =πππΈπΈ
=100 W
3.37 Γ 10β19 J
= 2.96 Γ 1020 photons/s
Thus, 2.96 Γ 1020 photons are delivered to the sphere every second.
8. The threshold frequency for a certain metal is 3.3 Γ 1014 π»π»π»π». If light of frequency 8.2 Γ 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Solution:
The threshold frequency of the metal is, ππ0 = 3.3 Γ 1014 Hz
The frequency of incident light on metal is, ππ = 8.2 Γ 1014 Hz
The charge of electron is, ππ = 1.6 Γ 10β19 C
The plankβs constant is, β = 6.626 Γ 10β34 Js
The cut off energy is,
ππ0 =β(ππ β ππ0)
ππ
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=6.626 Γ 10β34 Js Γ (8.2 Γ 1014 Hz β 3.3 Γ 1014 Hz)
1.6 Γ 10β19 C
= 2.03 V
Thus, the cut-off voltage for the photoelectric emission is 2.03 V.
9. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Solution:
Given
The work function is 4.2 eV.
The incident wavelength is ππ = 330 nm = 330 Γ 109 m
The energy of the incident photon is
πΈπΈ =βππππ
=6.626 Γ 10β34 Γ 3 Γ 108
330 Γ 109 m
= 6.0 Γ 10β19 J
=6.0 Γ 10β19
1.6 Γ 10β19= 3.76 eV
Here, as the incident radiation has an energy less than the work function of the metal. Thus, no photoelectric emission will take place.
10. Light of frequency 7.21 Γ 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 Γ 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Solution:
The frequency of the incident light is,
π£π£ = 488 nm = 488 Γ 10β9 m
The maximum speed of the electron is, π£π£ = 6.0 Γ 105 m/s
The plankβs constant is, 6.626 Γ 10β34 Js
The mass of the electron is, 9.1 Γ 10β31 kg
The kinetic energy is,
12πππ£π£
2 = β(ππ β ππ0)
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Here, ππ0 is the threshold frequency. Therefore, the threshold frequency is,
ππ0 = ππ βπππ£π£2
2β
= 7.21 Γ 1014 Hz β(9.1 Γ 10β31 kg)(6.0 Γ 105 m/s)2
2(6.626 Γ 10β34 Js)
= 7.21 Γ 1014 Hz β 2.47 Γ 1014 Hz
= 4.74 Γ 1014 Hz
Thus, 4.74 Γ 1014 Hz is the threshold frequency of the photoemission.
11. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Solution:
The wavelength of the light emitted from Argon laser is,
ππ = 488 nm = 488 Γ 10β9 m
The stopping potential of the photoelectron is,
ππ0 = 0.38 V
=0.38 V
1.6 Γ 10β19 V/eV
= 0.608 Γ 1019 eV
The Plankβs constant is, β = 6.626 Γ 10β34 Js
The charge of the electron is, ππ = 1.6 Γ 10β19 C
The speed of the electron is, ππ = 3 Γ 108 m/s
The work function is,
ππ =βππππβ ππππ0
οΏ½6.626 Γ 10β34 Js Γ 3 Γ 108 m/sοΏ½
οΏ½1.6 Γ 10β19 CοΏ½ οΏ½488 Γ 10β9 mοΏ½β οΏ½1.6 Γ 10β19 CοΏ½ οΏ½0.608 Γ 1019 eVοΏ½
= 2.16 eV
Thus, 2.16 eV is the work function required for the material.
12. Calculate the
(a) momentum, and
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(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Solution:
The potential difference is, V = 56 V
The Plankβs constant, β = 6.6 Γ 10β34 Js
The mass of electron is, ππ = 9.1 Γ 10β31 kg
The charge of the electron is, ππ = 1.6 Γ 10β19 C
(a) The kinetic energy of each electron at equilibrium is,
12πππ£π£
2 = ππππ
Here, π£π£ is the velocity.
Therefore, the velocity is,
π£π£ = οΏ½2ππππππ
= οΏ½2(1.6 Γ 10β19 C)(56 V)
(9.1 Γ 10β31 kg)
= 4.44 Γ 106 m/s
The momentum of each accelerated electron is,
ππ = πππ£π£
= (9.1 Γ 10β31 kg)(4.44 Γ 106 m/s)
= 4.04 Γ 10β24 kgm/s
Therefore, 4.04 Γ 10β24 kgm/s is the momentum of each electron.
(b) The de Broglie wavelength of the electron accelerated through a potential is,
ππ =12.27A
Β°
βππ
Here, ππ is the potential.
Therefore, de Broglie wavelength of electron is,
ππ =12.27 A
Β°
β56 V
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=οΏ½12.27 A
Β°οΏ½ οΏ½ 1 m
1010 mοΏ½
β56 V
=12.27 Γ 10β10 m
β56 V
= 0.16 Γ 10β9 m
= (0.16 Γ 10β9 m) οΏ½1 nm
10β9 mοΏ½
= 0.16 nm
Thus, 0.16 nm is the de Broglie wavelength of electron.
13. What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Solution:
Given
Kinetic energy of the electron is πΎπΎπΈπΈ = 120 eV.
(a) The speed of the electron is,
πΎπΎπΈπΈ =12πππ£π£2
π£π£ = οΏ½2(πΎπΎπΈπΈ)ππ
= οΏ½2 Γ 1.6 Γ 10β19 Γ 1209.1 Γ 10β31
= 6.496 Γ 106 m/s
The momentum of the electron is,
ππ = πππ£π£
= (9.1 Γ 10β31 kg)(6.496 Γ 106 m/s)
= 5.91 Γ 10β24 kgms
Thus, the momentum of the electron is 5.91 Γ 10β24 kg m/s.
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(b) The speed of the electron is 6.496 Γ 106 m/s.
(c) De Broglie wavelength of the electron is
ππ =βππ
=6.626 Γ 10β34
5.91 Γ 10β24
= 1.116 Γ 10β10 m
= 0.112 nm
Thus, the De Broglie wavelength of the electron is 0.112 nm.
14. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron would have the same de Broglie wavelength.
Solution:
The wavelength of the sodium light is,
ππ = 589 nm = 589 Γ 10β9 m
The mass of the electron is, ππππ = 9.1 Γ 10β31 kg
The mass of the neutron is, ππππ = 1.66 Γ 10β27 kg
The Plankβs constant is, β = 6.6 Γ 10β34 Js
(a) The kinetic energy of the electron is, πΎπΎ = 12πππππ£π£2
De Broglie wavelength of electron is, ππ = βπππππ£π£
Thus, the velocity of electron is, π£π£ = βππππππ
Thus, the kinetic energy of the electron is,
πΎπΎ =12ππππ οΏ½
βππππππ
οΏ½2
=β2
2ππ2ππππ
=(6.6 Γ 10β34 Js)2
2(589 Γ 10β9 m)(9.1 Γ 10β31 kg)
= 6.9 Γ 10β25 J
= (6.9 Γ 10β25 J) οΏ½1 eV
1.6 Γ 10β19 JοΏ½
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= 4.31 Γ 10β6 eV
= (4.31 Γ 10β6 eV) οΏ½1 ΞΌeV
10β6 eVοΏ½
= 4.31 ΞΌeV
The kinetic energy of electron is 4.31 ΞΌeV.
(b) The kinetic energy of neutron is,
πΎπΎ =β2
2ππ2ππππ
=(6.6 Γ 10β34 Js)2
2(589 Γ 10β9 m)(1.66 Γ 10β27 kg)
= 3.78 Γ 10β28 J
= (3.78 Γ 10β28 J) οΏ½1 eV
1.6 Γ 10β19 JοΏ½
= 2.36 Γ 10β9 eV
= (2.36 Γ 10β9 eV) οΏ½1 neV
10β9 eVοΏ½
= 2.36 neV
The kinetic energy of neutron is 2.36 neV.
15. What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 Γ 10β9 kg drifting with a speed of 2.2 m/s?
Solution:
(a) The mass of bullet is, ππ = 0.040 kg
The speed of bullet is,
π£π£ = 1 km
= (1 km) οΏ½1000 m
1 kmοΏ½
= 1000 m
The Plankβs constant is, β = 6.6 Γ 10β34 Js
The de Broglie wavelength of the bullet is,
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ππ =βπππ£π£
=6.6 Γ 10β34 J
(0.040 kg)(1000 m/s)
= 1.65 Γ 10β35 m
(b) The mass of ball is, ππ = 0.060 kg
The speed of ball is, π£π£ = 1.0 m/s
De Broglie wavelength of the ball is,
ππ =βπππ£π£
=6.6 Γ 10β34 J
(0.060 kg)(1 m/s)
= 1.1 Γ 10β32 m
(c) The mass of dust particle is, ππ = 1 Γ 10β9 kg
The speed of dust particle is, π£π£ = 2.2 m/s
The de Broglie wavelength of the dust particle is,
ππ =βπππ£π£
=6.6 Γ 10β34 J
(1 Γ 10β9 kg)(2.2 m/s)
= 3.0 Γ 10β25 m
16. An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Solution:
The wavelength of the photon is ππππ = 1.00 nm = 1 Γ 10β9 m.
The wavelength of the electron is ππππ = 1.00 nm = 1 Γ 10β9 m.
(a) According to De Broglie relation, the momentum is
ππ =βππ
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As the momentum depends only on the wavelength, both photon and electron have the same momentum.
ππ =6.626 Γ 10β34
1 Γ 10β9
= 6.626 Γ 10β25 kgm/s
The momentum of photon and electron is 6.626 Γ 10β25 kgm/s.
(b) The energy of photon is given by,
πΈπΈ =βππππ
=6.626 Γ 10β34 Γ 3 Γ 108
1 Γ 10β9 Γ 1.6 Γ 10β19
= 1243.1 eV
= 1.243 keV
(c) The kinetic energy of the electron is,
πΎπΎ =12ππ2
ππ
=12
(6.63 Γ 10β25)2
9.1 Γ 10β31 Γ 1.6 Γ 10β19
=2.415 Γ 10β19 J
1.6 Γ 10β19
= 1.51 eV
17. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 Γ 10β10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Solution:
(a) De Broglie wavelength of the neutron is 1.40 Γ 10β10 m.
Neutron has a mass of ππππ = 1.66 Γ 10β27 kg.
The relation between kinetic energy and velocity is,
πΎπΎ = 12πππππ£π£2 β¦β¦ (i)
The relation between De Broglie wavelength and velocity is,
ππ = βπππππ£π£
β¦β¦ (ii)
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Comparing equations (i) and (ii),
πΎπΎ =12ππππβ2
ππ2ππππ2 =
β2
2ππ2ππππ
Substituting the values,
πΎπΎ =(6.626 Γ 10β34)2
2(1.40 Γ 10β10)2 Γ 1.66 Γ 10β27
= 6.75 Γ 10β21 J
=6.75 Γ 10β21
1.6 Γ 10β19
= 4.219 Γ 10β2 eV
(b) Average kinetic energy of the neutron is,
πΎπΎβ² =32ππππ
=32
Γ 1.38 Γ 10β23 Γ 300
= 6.21 Γ 10β21 J
The De Broglie wavelength is,
ππ =β
οΏ½2πΎπΎβ²ππππ
=6.626 Γ 10β34
β2 Γ 6.21 Γ 10β21 Γ 1.66 Γ 10β27
= 1.46 Γ 10β10 m
= 0.146 mm
18. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
The momentum of the photon is,
ππ =πΈπΈππ
=βππππ
=βππ
On rearranging, ππ = βππ β¦β¦ (i)
The De Broglie wavelength is, ππ = βπππ£π£
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Substitute ππfor πππ£π£.
ππ = βππ β¦β¦ (ii)
As equation (i) and (ii) are the same, e wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum.
19. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Solution:
Given
The atomic mass of nitrogen is 14.0076 u.
Thus, the mass of nitrogen molecule is,
ππ = 2 Γ 14.0076 u
= 28.0152 u
= 28.0152 Γ 1.66 Γ 10β27
The relation between the kinetic energy of the nitrogen molecule and the root mean square speed is,
12πππ£π£ππππππ
2 =32ππππ
π£π£ππππππ = οΏ½3ππππππ
Thus, the De Broglie wavelength of the nitrogen molecule is,
ππ =β
πππ£π£ππππππ=
ββ3ππππππ
=6.626 Γ 10β34
β3 Γ 28.0152 Γ 1.66 Γ 10β27 Γ 1.38 Γ 10β23 Γ 300
= 0.028 Γ 10β9 m
= 0.028 m
Thus, the De Broglie wavelength of the nitrogen molecule is, 0.028 m.
20. (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its ππ/ππ is given to be 1.76 Γ 1011 C kgβ1.
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(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?
Solution:
(a) Given
The potential difference across the tube is, ππ = 500 V.
The specific charge of the electron is, ππ/ππ = 1.76 Γ 1011 C/kg.
The speed of the emitted electron is,
K =12πππ£π£2 = ππππ
π£π£ = οΏ½2ππππππ
= οΏ½2ππππππ
Substituting the values,
π£π£ = οΏ½2 Γ 500 Γ 1.76 Γ 1011
= 1.327 Γ 107 m/s
Thus, the speed of the emitted electron is 1.327 Γ 107 m/s.
(b) Given
The potential of the anode is, ππ = 10 MV = 10 Γ 106 V.
The speed of each electron is, π£π£ = οΏ½2ππ ππππ
.
Substituting the values,
π£π£ = οΏ½2 Γ 107 Γ 1.76 Γ 1011
= 1.88 Γ 109 m/s
This speed is greater than the speed of light which is not possible. This is because the expression for energy πππ£π£
2
2 can be used only in non-relativistic situations (π£π£ << ππ). In relativistic situations, the total energy is written as
πΈπΈ = ππππ2
Here, ππis the relativistic mass given by ππ = ππ0οΏ½1 β π£π£2
ππ2.
Here, ππ0is the mass of the particle at rest.
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Thus, the kinetic energy can be written as
πΎπΎ = ππππ2 β ππ0ππ2
21. (a) A monoenergetic electron beam with electron speed of 5.20 Γ 106 m sβ1 is subject to a magnetic field of 1.30 Γ 10β4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 Γ 1011C kgβ1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of
Solution:
(a) The speed of the electron is, π£π£ = 5.20 Γ 106 m/s
The magnetic field is, π΅π΅ = 1.30 Γ 10β4 T
The specific charge of the electron is,
ππππ =
1.6 Γ 10β19 C9.1 Γ 10β31 kg
= 1.76 Γ 1011 C/kg
The force exerted on the electron is,
πΉπΉ = πποΏ½οΏ½βοΏ½π£ Γ π΅π΅οΏ½β οΏ½
= πππ£π£π΅π΅sinππ
Here, ππ is the angle between velocity and magnetic field. Since the magnetic field is normal to the direction of beam, the force is,
πΉπΉ = πππ£π£π΅π΅sin90Β° = πππ£π£π΅π΅
The magnetic force acting on the electron will allow it to move in a circular path. Centripetal force acting on the electron is, πΉπΉ = πππ£π£2
ππ
The magnetic force is equal in magnitude with the centripetal force. Therefore,
πππ£π£π΅π΅ =πππ£π£2
ππ
Here, ππ is the radius of circular path of the electron. Thus, the radius of electron is,
ππ =πππ£π£πππ΅π΅
=π£π£
(ππ/ππ)π΅π΅
=5.20 Γ 106 m/s
(1.76 Γ 1011 C/kg)(1.30 Γ 10β4 T)
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= 0.227 m or 22.7 cm
Thus, 22.7 cm is the radius of the circular path.
(b) The energy of the electron beam is,
πΈπΈ = 20MeV
= (20MeV) οΏ½1 eV
10β6 MeVοΏ½
= 20 Γ 106 eV
= (20 Γ 106 eV)οΏ½1.6 Γ 10β19 J
1 eVοΏ½
= 32 Γ 10β13 J
Kinetic energy of the electron is,
πΈπΈ =12πππ£π£2
Therefore, the speed of the electron is,
π£π£ = οΏ½2πΈπΈππ
= οΏ½2 Γ 36 Γ 10β13 J9.1 Γ 10β31 kg
= 2.65 Γ 109 m/s
No objects can move faster than light. Therefore, the result is incorrect. Therefore, the energy equation can only use in non-relativistic case where π£π£ << ππ. In relativistic case, mass is given by,
ππ = ππ0 οΏ½1 βπ£π£2
ππ2οΏ½1/2
Here, ππ0 is the rest mass of particle.
Therefore, radius of the circular path is,
ππ = πππ£π£ππππ
= ππ0π£π£
πππποΏ½ππ2βπ£π£2
ππ2
22. An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~10β2 mm of Hg). A magnetic field of 2.83 Γ 10β4 T curves the path of the electrons in a circular orbit
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of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons and emitting light by electron capture; this method is known as the βfine beam tubeβ method.) Determine e/m from the data.
Solution:
The potential of the anode is, ππ = 100 V
The magnetic field is, π΅π΅ = 2.83 Γ 10β4T
The radius of the circular orbit is,
ππ = 12.0 cm
= (12.0 cm) οΏ½1 m
100 cmοΏ½
= 12.0 Γ 10β2 m
The kinetic energy of the electron and the electronβs energy is equal in magnitude. Therefore,
12πππ£π£
2 = ππππ
Here, ππ is the mass of electron, ππ is the charge of electron and π£π£ is the
Therefore,
π£π£2 =2ππππππ
The centripetal force and the magnetic force acting on the electron is equal in magnitude. Therefore,
πππ£π£2
ππ = πππ£π£π΅π΅
β πππ΅π΅ =πππ£π£ππ
β π£π£ =πππ΅π΅ππππ
Therefore,
2ππππππ = οΏ½
πππ΅π΅ππππ οΏ½
2
Therefore,
ππππ =
2πππ΅π΅2ππ2
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=2 Γ (100 V)
(2.83 Γ 10β4 T)2 Γ (12.0 Γ 10β2 m)2
= 1.73 Γ 1011 C/kg
Thus, 1.73 Γ 1011 C/kg is the specific charge ratio.
23. (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Γ . What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Solution:
(a) Given,
The wavelength of the X-ray is, ππ = 0.45 Γ = 0.45 Γ 10β10 m.
The maximum energy of the photon in the radiation is,
πΈπΈ =βππππ
=6.626 Γ 10β34 Γ 3 Γ 108
0.45 Γ 10β10 Γ 1.6 Γ 1019
= 27.6 Γ 103 eV
= 27.6 keV
Thus, the maximum energy of the photon in the radiation is 27.6 keV.
(b) The photons have a maximum energy of 27.6 keV. The energy for producing X-rays by electrons is given by accelerating voltage. For the X-ray to have an energy of 27.6 keV, the electrons must have at least an energy of 27.6 keV. Thus, for producing X-rays, an accelerating voltage of the order of 30 keVis required.
24. In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two πΎπΎ-rays of equal energy. What is the wavelength associated with each πΎπΎ-ray? (1 BeV = 109 eV)
Solution:
Given,
The total energy of two πΎπΎ-rays is
πΈπΈ = 10.2 BeV
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= 10.2 Γ 109 eV
= 10.2 Γ 109 Γ 1.6 Γ 10β19 J
The energy of each πΎπΎ-ray is,
πΈπΈβ² =πΈπΈ2
=10.2 Γ 109 Γ 1.6 Γ 10β19 J
2
= 8.16 Γ 10β10 J
The wavelength associated with πΎπΎ-ray is,
πΈπΈβ² =βππππ
ππ =βπππΈπΈβ²
=6.626 Γ 10β34 Γ 3 Γ 108
8.16 Γ 10β10
= 2.436 Γ 10β16 m
Thus, the wavelength associated with πΎπΎ-ray is 2.436 Γ 10β16 m.
25. Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never βcount photonsβ, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~10β 10 W mβ2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 Γ 1014 Hz.
Solution:
(a) Given
The wave transmitterβs power is ππ = 10 kW = 104 W = 104 J/s.
Thus, the energy emitted per second by the transmitter is 104 J.
The radio wave has a wavelength of ππ = 500 m.
The energy of the wave is
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πΈπΈ1 =βππππ
=6.6 Γ 10β34 Γ 3 Γ 108
500
= 3.96 Γ 10β28 J
Assume the n is the number of photons that the transmitter emits.
Thus,
πππΈπΈ1 = πΈπΈ
ππ =πΈπΈπΈπΈ1
=104
3.96 Γ 10β28
= 2.525 Γ 1031
β 3 Γ 1031
Even though the number of photons emitted per second is large, the energy of the emitted photon is less.
(b) Given
The minimum intensity of light perceived by the human eye is, πΌπΌ =10β10 W/m2.
The average frequency of the white light is, ππ = 6 Γ 1014 Hz.
The area of the pupil is, π΄π΄ = 0.4 cm2 = 0.4 Γ 10β4 m2.
The energy emitted by the photon is,
πΈπΈ = βππ
= 6.626 Γ 10β34 Γ 6 Γ 1014
= 3.96 Γ 10β19 J
Let ππ be a number that represents the number of photons falling per second per unit area of the pupil.
For n falling protons, the total energy per unit area per unit second is,
πΈπΈ = ππ Γ 3.96 Γ 10β19 J sβ1mβ2
As intensity is the energy per unit area per second,
πΈπΈ = πΌπΌ
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ππ Γ 3.96 Γ 10β19 = 10β10
ππ =10β10
3.96 Γ 10β19
= 2.52 Γ 108 m2 sβ1
Total number of photons entering the pupil per second is given by
πππ΄π΄ = πππ΄π΄
= 2.52 Γ 108 Γ 0.4 Γ 10β4
= 1.008 Γ 104 ππβ1
Even though the number is not as large as the number in (a), it is large enough that the human eye cannot see individual photons.
26. Ultraviolet light of wavelength 2271 Γ from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is β 1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~105 W mβ2) red light of wavelength 6328 Γ produced by a He-Ne laser?
Solution:
Given
The wavelength of the ultraviolet light is, ππ = 2271 Γ = 2271 Γ 10β10 m.
The metal has a stopping potential of ππ0 = 1.3 V.
The equation for work function according to photoelectric effect is,
ππ0 = βππ β ππππ0
=βππππβ ππππ0
=6.626 Γ 10β34 Γ 3 Γ 108
2271 Γ 10β10β 1.6 Γ 10β19 Γ 1.3
= 8.72 Γ 10β19 β 2.08 Γ 10β19
= 6.64 Γ 10β19 J
=6.64 Γ 10β19
1.6 Γ 10β19
= 4.15 eV
If ππ0is the threshold frequency of the metal, then
ππ0 = βππ0
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β ππ0 =ππ0β
=6.64 Γ 10β19
6.6 Γ 10β34
= 1.006 Γ 1015 Hz
The frequency of the red light is, ππππ = ππππππ
.
Substitute 6328 Γ 10β10 mfor ππππin the above equation,
ππππ =3 Γ 108
6328 Γ 10β10 = 4.74 Γ 1014 Hz
Thus, the photocell will not respond to the red light from the laser as ππ0 > ππππ.
27. Monochromatic radiation of wavelength 640.2 nm (1 nm = 10β9 m) from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage.
Solution:
Given
Wavelength of the monochromatic radiation is, ππ = 640.2 nm = 640.2 Γ 10β9 m.
The stopping voltage for the lamp is, ππ0 = 0.54 V.
According to photoelectric effect, the relation between work function and frequency is,
ππππ0 = βππ β ππ0
ππ0 =βππππ β ππππ0
=6.626 Γ 10β34 Γ 3 Γ 108
640.2 Γ 10β9β 1.6 Γ 10β19 Γ 0.54
= 3.093 Γ 10β19 β 0.864 Γ 10β19
= 2.229 Γ 10β19 J
=2.229 Γ 10β19
1.6 Γ 10β19
= 1.39 eV
The wavelength of the wave from the iron source isππβ² = 427.2 nm =427.2 Γ 10β9 m.
From the photoelectric equation, the new stopping potential is,
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ππππ0β² =βππππβ ππ0
=6.626 Γ 10β34 Γ 3 Γ 108
427.2 Γ 10β9β 2.229 Γ 10β19
= 4.63 Γ 10β19 β 2.229 Γ 10β19
= 2.401 Γ 10β19 J
=2.401 Γ 10β19
1.6 Γ 10β19
= 1.5 eV
Thus, the new stopping potential is 1.5 eV.
28. A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used: ππ1 = 3650 Γ , ππ2 = 4047 Γ , ππ3 = 4358 Γ , ππ4 = 5461 Γ , ππ5 = 6907 Γ , The stopping voltages, respectively, were measured to be: ππ01 = 1.28 ππ, ππ02 = 0.95 ππ, ππ03 = 0.74 ππ, ππ04 = 0.16 ππ, ππ05 = 0 ππ. Determine the value of Planckβs constant β, the threshold frequency and work function for the material.
Solution:
According to Einsteinβs photoelectric equation,
ππππ0 = βππ β ππ0
β ππ0 =βππππ β
ππ0ππ
Here, ππ0 is the stopping potential, β is the Plankβs constant, ππ is the charge of electron, ππ is the frequency of the radiation and ππ0 is the work function of a material.
The stopping potential is directionally proportional to frequency.
The frequency is, ππ = ππππ
Here, ππ is the speed of light and ππ is the wavelength.
Therefore, the frequencies for various lines of given wavelengths can be obtained.
ππ1 =ππππ1
=3 Γ 108 m/s
3650 Γ 10β10 m= 8.219 Γ 1014 Hz
ππ2 =ππππ2
=3 Γ 108 m/s
4047 Γ 10β10 m= 7.412 Γ 1014 Hz
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ππ3 =ππππ3
=3 Γ 108 m/s
4358 Γ 10β10 m= 6.884 Γ 1014 Hz
ππ4 =ππππ4
=3 Γ 108 m/s
5461 Γ 10β10 m= 5.493 Γ 1014 Hz
ππ5 =ππππ5
=3 Γ 108 m/s
6907 Γ 10β10 m= 4.343 Γ 1014 Hz
The tabular form of the data is,
Frequency Γ 1014 Hz 8.219 7.412 6.884 5.493 4.343
StoppingPotentialππ0 1.28 0.95 0.74 0.16 0
The graph between frequency and stopping potential is,
The curve of the graph is straight line and it intersects the axis of frequency at 5 Γ 1014 Hz. Therefore, 5 Γ 1014 Hz is the threshold frequency of the material. Therefore, for ππ5 there will not be any photo electric emission and no stopping voltage is required.
Slope of the line is,
π΄π΄π΅π΅πΆπΆπ΅π΅ =
1.28β 0.16 V(8.214β 5.493) Γ 1014 Hz
= 4.12 Γ 10β15 V/Hz
From equation for stopping potential, the slope is,
βππ = 4.12 Γ 1015 V/Hz
Therefore, the Plankβs constant is,
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β = (4.12 Γ 1015 V/Hz)ππ
= (4.12 Γ 1015 V/Hz)(1.6 Γ 10β19 C)
= 6.573 Γ 10β34 Js
The work function of the metal is,
ππ0 = βπ£π£0
= (6.573 Γ 10β34 Js)(5 Γ 1014 Hz)
= 3.286 Γ 10β19 J
= (3.286 Γ 10β19 J) οΏ½1 eV
1.6 Γ 10β19 JοΏ½
= 2.054 eV
29. The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Γ from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and place 50 cm away?
Solution:
The wavelength of the radiation is,
ππ = 3300 Γ
= (3300 Γ ) οΏ½1 m
1010 Γ οΏ½
= 3300 Γ 10β10 m
The speed of light is, ππ = 3 Γ 108 m/s
The Plankβs constant is, β = 6.6 Γ 10β34 Js
The energy of the incident radiation is,
πΈπΈ =βππππ
=6.6 Γ 10β34 J Γ 3 Γ 108 m/s
3300 Γ 10β10 m
= 6 Γ 10β19 J
= (6 Γ 10β19 J) οΏ½1 eV
1.6 Γ 10β19 JοΏ½
= 3.16 eV
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The energy exceeds the work function of K and Na alone. Therefore, both Ni and Mo will not show photo electric emission. The intensity of incident light increases as the source brought near. However, it will not affect the energy of the radiation, but the photoelectron emitted from K and Na increase in proportion to intensity.
30. Light of intensity 10β5 W mβ2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:
The intensity of the incident light is, πΌπΌ = 10β5 W/m2
The surface area of the sodium cell is,
π΄π΄ = 2 cm2
= (2 cm2)οΏ½1 m2
104 cm2οΏ½
= 2 Γ 10β4 m2
The power if the incident life is,
ππ = πΌπΌπ΄π΄
= 10β5 W/m2 Γ 2 Γ 10β4 m2
= 2 Γ 10β9 W
The work function of the metal,
ππ0 = 2 eV
= (2 eV) οΏ½1.6 Γ 10β19 J
1 eVοΏ½
= 3.2 Γ 10β19 J
The number of layers of sodium that absorbs the incident energy, ππ = 5
The effective atomic area of sodium atom is, π΄π΄ππ = 10β20 m2
Therefore, the number of conduction electrons in ππ layers is,
ππβ² = ππ Γπ΄π΄π΄π΄ππ
= 5 Γ2 Γ 10β4 m2
10β20 m2
= 1017
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The electrons uniformly absorb incident power. Amount of energy absorbed per second per electron is given by,
πΈπΈ =ππππβ²
=2 Γ 10β9 W
1017
= 2 Γ 10β26 J/s
The time required for photoelectric emission is,
π‘π‘ =ππ0πΈπΈ
=3.2 Γ 10β19
2 Γ 10β26
= 1.6 Γ 107 s
= (1.6 Γ 107 s) οΏ½1 year
3.15 Γ 107 sοΏ½
= 0.51 years
The photoelectric emission will not take half a year. Therefore, the wave picture is not incorrect.
31. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Γ , which is of the order of inter-atomic spacing in the lattice) (me = 9.11 Γ 10β31 kg).
Solution:
Given
The wavelength of the probe is, ππ = 1 Γ = 1 Γ 10β10 m.
The mass of the electron is, ππππ = 9.1 Γ 10β31 kg.
For the same wavelength, an X-ray probe has greater energy than an electron probe.
The kinetic energy of the electron is,
πΈπΈ =12πππππ£π£2
β π£π£ = οΏ½2πΈπΈππππ
πππππ£π£ = οΏ½2πΈπΈππππ
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Here, πππππ£π£ is the momentum of the electron.
The De Broglie wavelength, according to De Broglie relation is,
ππ =βππ
=β
πππππ£π£=
βοΏ½2πΈπΈππππ
πΈπΈ =β2
2ππ2ππππ
=(6.626 Γ 10β34)2
2(1 Γ 10β10)2 Γ 9.11 Γ 10β31
= 2.39 Γ 10β17 J
=2.39 Γ 10β17
1.6 Γ 10β19
= 12.375 Γ 103eV
= 12.375 keV
Thus, it can be concluded that, for the same wavelength, a proton has greater than an electron.
32. (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 Γ 10β27 kg)
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 Β°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Solution:
(a) The de Broglie wavelength is, 2.327 Γ 10β12 m
The kinetic energy of neutron is,
πΎπΎ = 150 eV
= (150 eV)οΏ½1.6 Γ 10β19 J
1 eVοΏ½
= 2.4 Γ 10β17 J
The mass of the neutron is, ππππ = 1.675 Γ 10β27 kg
The kinetic energy of the neutron is,
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πΎπΎ =12πππππ£π£2
Here, π£π£ is the velocity of neutron.
Therefore, the momentum of the neutron is,
πππππ£π£ = οΏ½2πΎπΎππππ
De Broglie wavelength of the neutron is,
ππ =β
πππππ£π£=
βοΏ½2πΎπΎππππ
Therefore, the wavelength is inversely proportional to the square root of the mass.
The wavelength is,
ππ =6.6 Γ 10β34 J
οΏ½2 Γ 2.4 Γ 10β17 J Γ 1.675 Γ 10β27 kg
= 2.327 Γ 10β12 m
The inter atomic spacing of a crystal is about 10β10 m. Thus, 150 eVneutron beam is not suitable for diffraction.
(b) The de Broglie wavelength is, 1.44 Γ 10β10 m
The Boltzmann constant is 1.38 Γ 10β23 J/molK
The room temperature is,
ππ = 27Β°C
= (27 + 273) K
= 300 K
The average kinetic energy of neutron is,
πΈπΈ =32ππππ
The wavelength of the neutron is,
ππ =β
οΏ½2πππππΈπΈ=
βοΏ½3ππππππππ
=6.6 Γ 10β34 Js
οΏ½3(1.675 Γ 10β27 kg)(1.38 Γ 10β23 J/molK)(300 K)
= 1.447 Γ 10β10 m
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Therefore, the neutron beam can be used to diffraction after thermalizing.
33. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Solution:
Given
The voltage used to accelerate the electrons is, ππ = 50 kV = 50 Γ 103 V.
The yellow light has a wavelength of, ππ = 5.9 Γ 10β7 m.
The kinetic energy of the electron is,
πΈπΈ = ππππ
= 1.6 Γ 10β19 Γ 50 Γ 103
= 8 Γ 10β15 J
According to De Broglie relation, the wavelength is,
ππ =β
οΏ½2πππππΈπΈ
=6.6 Γ 10β34
β2 Γ 9.11 Γ 10β31 Γ 8 Γ 10β15
= 5.467 Γ 10β12 m
The relation between wavelength and resolving power of a microscope is that the resolving power of a microscope depends inversely on the wavelength of the light used. The De Broglie wavelength is about 105times smaller than the wavelength of the yellow light. Thus, an electron microscope has 105times the resolving power compared to an optical microscope.
34. The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10β15 m or less. This structure was first probed in early 1970βs using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Solution:
The wavelength of the proton or a neutron, ππ β 10β15 m
The rest mass energy of an electron is,
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ππ0ππ2 = 0.511 MeV
= (0.511 MeV)οΏ½106 eV1 MeV
οΏ½οΏ½1.6 Γ 10β19 J
1 eVοΏ½
= 0.8176 Γ 10β13 J
The Plankβs constant is, β = 6.6 Γ 10β34 Js
The speed of light, ππ = 3 Γ 108 m/s
The momentum of a proton or a neutron is,
ππ =βππ
=6.6 Γ 10β34 Js
10β15 m
= 6.6 Γ 10β19 kg m/s
The relativistic equation for energy is,
πΈπΈ2 = ππ2ππ2 +ππ02ππ4
β πΈπΈ = οΏ½ππ2ππ2 + ππ02ππ4
= οΏ½(6.6 Γ 10β19 kgm/s Γ 3 Γ 108 m/s)2 + (0.8176 Γ 10β13)2
= οΏ½392 Γ 10β22 J2
= 1.98 Γ 10β10 J
= (1.98 Γ 10β10 J) οΏ½1 eV
1.6 Γ 10β19 JοΏ½
= 1.24 Γ 109 eV
= (1.24 Γ 109 eV) οΏ½1 BeV109 eV
οΏ½
= 1.24 BeV
Therefore, 1.24 BeVelectron energy is emitted.
35. Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 Β°C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Solution:
De Broglie wavelength associated with He atom is, 0.7268 Γ 10β10 m
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The room temperature is,
ππ = 27 Β°C
= (27 + 273) K
= 300 K
The atmospheric pressure is,
ππ = 1 atm
= (1 atm)οΏ½1.01 Γ 105 Pa
1 atmοΏ½
= 1.01 Γ 105 Pa
The atomic weight of a He atom ππ = 4 g
The Avogadroβs number is, πππ΄π΄ = 6.023 Γ 1023
The Boltzmann constant is, ππ = 1.38 Γ 10β23 J/molK
The average energy of a gas is,
πΈπΈ =32ππππ
Here, ππ is the temperature.
The de Broglie wavelength is,
ππ =β
β2πππΈπΈ
Here, ππis the mass of He atom.
ππ =πππππ΄π΄
=4 g
6.023 Γ 1023
= 6.64 Γ 10β24 g
= (6.64 Γ 10β24 g) οΏ½1 kg
1000 gοΏ½
= 6.64 Γ 10β27 g
The de Broglie wavelength is,
ππ =β
β3ππππππ
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=6.6 Γ 10β34 Js
οΏ½(3)(6.64 Γ 10β27 kg)(1.38 Γ 10β23 J/mol K) Γ 300 K
= 0.7268 Γ 10β10 m
The ideal gas formula is,
ππππ = π π ππ
ππππ = ππππππ
ππππ =
ππππππ
Here, ππ is the volume of the gas and ππ is the number of moles of the gas.
The mean separation between two atoms of the gas is,
ππ = οΏ½πππποΏ½1/3
= οΏ½πππππποΏ½1/3
= οΏ½(1.38 Γ 10β23 J/molK)(300 K)
1.01 Γ 105 PaοΏ½
= 3.35 Γ 10β9 m
Therefore, the mean separation between two atoms of the gas is greater than de Broglie wavelength.
36. Compute the typical de Broglie wavelength of an electron in a metal at 27 Β°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 Γ 10β10 m.
Solution:
The Plankβs constant is, β = 6.6 Γ 10β34 Js
The mass of the electron is, ππ = 9.11 Γ 10β31 kg
The Boltzmann constant is, ππ = 1.38 Γ 10β23 J/molK
The room temperature is,
ππ = 27Β°C
= (27 + 273) K
= 300 K
The mean separation between two electrons is, ππ = 2 Γ 10β10 m
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The de Broglie wavelength is,
ππ =β
β3ππππππ
=6.6 Γ 10β34 J s
οΏ½2(9.11 Γ 10β31 kg)(1.38 Γ 10β23 J/molK)(300 K)
= 6.2 Γ 10β9 m
Therefore, the inter-electron separation is much less than de Broglie wavelength.
37. Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)ππ ; (β 1/3)ππ]. Why do they not show up in Millikanβs oil-drop experiment?
(b) What is so special about the combination ππ/ππ? Why do we not simply talk of ππ and ππ separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: πΈπΈ = βππ,ππ = β
ππ.
But while the value of ππ is physically significant, the value of ππ (and therefore, the value of the phase speed ππππ) has no physical significance. Why?
Solution:
(a) The quarks situated inside neutron and proton carries fractional charges since if the pulled apart, the nuclear force increase extremely. Therefore, fractional charges can be present in nature and the observable charges are the integral multiple of electric charge.
(b) The kinetic energy of an electron moving in a potential is,
ππππ =12πππ£π£2
Here, ππ is the charge of electron, ππ is the potential, ππis the mass of electron and π£π£ is the velocity of electron.
The velocity is,
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π£π£ = οΏ½2ππ οΏ½πππποΏ½
The centripetal force acting on an electron in a magnetic field is,
πππ΅π΅π£π£ =πππ£π£2
ππ
Here, π΅π΅ is the magnetic field and ππ is the radius of rotation of the electron.
The velocity is,
π£π£ = π΅π΅ππ οΏ½πππποΏ½
Therefore, the dynamics of the electron is determined by the ratio ππ/ππ and not by ππ and ππ separately.
(c) Gases are insulator at atmospheric pressure since the ions of gases will not have any chance to reach their respective electrons at due to the collision. The ions have a chance to reach their respective electrons and produce an electric current only at low pressure.
(d) The work function of a metal is defined as the minimum energy required for an electron to get out of the surface of the metal. The electrons in an atom will have different energy levels. When a photon ray of some energy is incident on the metal, electrons of different energies are emitted from different levels. Thus, the photoelectrons show different energy distributions.
(e) As the absolute value of energy for a particle is arbitrary within the additive constant, the frequency associated with an electron has no physical significance whereas the wavelength is significant.
Thus, the product of frequency and wavelength ππππhas no physical significance.
The group speed can be shown as
π£π£G =ππππππππ
=ππππ
ππ οΏ½1πποΏ½
=πππΈπΈππππ
=ππ οΏ½ ππ
2
2πποΏ½
ππππ=ππππ
This is a quantity with physical meaning.
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