Ncert Solution Class 12th Physics Chapter – 11 Dual Nature of Radiation and Matter
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 11 |
| Chapter Name | Dual Nature of Radiation and Matter |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 11 Dual Nature of Radiation and Matter
?Chapter – 11?
✍Dual Nature of Radiation and Matter✍
?Question & Answer?
Q 1. Find the:
(a) Maximum frequency, and
(b) The minimum wavelength of X-rays produced by 30 kV electrons.
Answer: Potential of the electrons, V = 30 kV = 3 × 104 V
Hence, energy of the electrons, E = 3 × 104 eV
Where,
e = Charge on an electron = 1.6 × 10−19 C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E = hν
Where,
h = Planck’s constant = 6.626 × 10−34 Js
So v = E / h
= 1.6 x 10-19 x 3 x 104 / 6.626 x 10-34 = 7.24 x 1018 Hz
Hence, the maximum frequency of X-rays produced is 7.24 x 1018 Hz
(b)The minimum wavelength produced by the X-rays is given as:
λ = c / v
= 3 x 108 / 7.24 x 1018 = 4.14 x 10-11 m = 0.0414 nm
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
Q 2. The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons
(b) Stopping potential
(c) maximum speed of the emitted photoelectrons?
Answer: Work function of caesium metal, ø0 = 2.14 eV
Frequency of light, v = 6.0 x 1014 Hz
(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv – ø0
Where,
h = Planck’s constant = 6.626 × 10−34 Js
∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 – 2.14 = 2.485 – 2.14 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0
∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V
Hence, the stopping potential of the material is 0.345 V.
(c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as: K = 1/2 mv2
Where,
m = Mass of an electron = 9.1 × 10−31 kg
v2 = 2K/m = 2 x 0.345 x 1.6 x 10-19 / 9.1 x 10-31 = 0.1104 x 1012
∴ v = 3.323 x 105 m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?
Answer: Photoelectric cut-off voltage, Vo = 1.5 V
For emitted photoelectrons, the maximum kinetic energy is:
Ke = eVo
Where,
e = charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J
Therefore, 2.4 x 10-19 J is the maximum kinetic energy emitted by the photoelectrons.
Q 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 a uniform cross-section which is less than the target area)
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer: Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg
(a) The energy of each photon is given as: E = hc/λ = 3.141 x 10-19 J
The momentum of each photon is given as: P = h/λ = 6.626 x 10-34 / 632.8 = 1.047 x 10-27 kg m s-1
(b) Number of photons arriving per second, at a target irradiated by the beam = n Assume that the beam has a uniform cross-section that is less than the target area. Hence, the equation for power can be written as:
P= nE
∴ n = P/E = 9.42 x 10-3 / 3.141 x 10-19 = 3 x 1016 photons/s (approx)
(c) Momentum of the hydrogen atom is the same as the momentum of the photon,
p = 1.047 x 10-27 kg m s-1
Momentum is given as: p = mv
Where,
v = Speed of the hydrogen atom
∴ v = p/m = 1.047 x 10-27 / 1.66 x 10-27 = 0.621 m/s
Q 5. The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons are incident on the Earth per second/square meter? Assume an average wavelength of 550 nm.
Answer: Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 103 W/m2
Hence, power of sunlight per square metre, P = 1.388 × 103 W
Speed of light, c = 3 × 108 m/s
Planck’s constant, h = 6.626 × 10−34 Js
Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10-9 m
Number of photons per square metre incident on earth per second = n Hence, the equation for power can be written as: P= nE
∴ n = P/E = Pλ / hc = 1.388 x 103 x 550 x 10-9 / 6.626 x 10-34 x 3 x 108 = 3.84 x 1021 photons/m2/s
Therefore, every second, 3.84 x 1021 photons are incident per square metre on earth.
Q 6. In an experiment on the photoelectric effect, the slope of the cut-off
voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.
Answer: The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as: V/v = 4.12 x 10-15 VsV is related to frequency by the equation: hv = eV
Where,
e = Charge on an electron = 1.6 × 10−19 C
h = Planck’s constant
∴ h = e x V/v = 1.6 x 10-19 x 4.12 x 10-15 = 6.592 x 10-34 Js
Therefore, the value of Planck’s constant is 6.592 x 10-34 Js
Q 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?
Answer: Power of the sodium lamp, P = 100 WWavelength of the emitted sodium light, λ = 589 nm = 589 × 10−9 m
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
(a) The energy per photon associated with the sodium light is given as: E = hc/λ = 6.626 x 10-34 x 3 x 108 / 589 x 10-9 = 3.37 x 10-19 J
= 3.37 x 10-19 / 1.6 x 10-19 = 2.11 eV
(b) Number of photons delivered to the sphere = n.
The equation for power can be written as: P = nE
∴ n = P/E = 100 / 3.37 x 10-19 = 2.96 x 1020 photons/s
Therefore, every second, 2.96 x 1020 photons are delivered to the sphere.
Q 8. The threshold frequency for a certain metal is 3.3 x 1014 Hz. If the light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer: Threshold frequency of the metal, V0 = 3.3 x 1014 Hz
Frequency of light incident on the metal, V = 8.2 x 1014 Hz
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.626 × 10−34 Js
Cut-off voltage for the photoelectric emission from the metal = V0
The equation for the cut-off energy is given as: eV0 = h (v – v0)
∴ v0 = h(v – v0) / e = 6.626 x 10-34 x (8.2 x 1014 – 3.3 x 1014) / 1.6 x 10-19 = 2.0292 V
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V
Q 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?
Answer: No Work function of the metal, ø0 = 4.2 eV
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.626 × 10−34 Js
Wavelength of the incident radiation, λ = 330 nm = 330 × 10−9 m
Speed of light, c = 3 × 108 m/s
The energy of the incident photon is given as: E = hc/λ
= 6.626 x 10-34 x 3 x 108 / 330 x 10-9 = 6.0 x 10-19 J = 6.0 x 10-19 / 1.6 x 10-19 = 3.76 eV
It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.
Q 10. Light of frequency 7.21 x 1014 Hz is incident in a metal surface. Electrons with a maximum speed of 6.0 x 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer: Frequency of the incident photon, v = 488 nm = 488 x 10-9 m
Maximum speed of the electrons, v = 6.0 × 105 m/s
Planck’s constant, h = 6.626 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
For threshold frequency ν0, the relation for kinetic energy is written as: 1/2 mv2
= h(v – v0)
∴ v0 = v – mv2/2h
= 7.21 x 1014 – (9.1 x 10-31) x (6 x 105)2 / 2 x (6.626 x 10-34)
= 7.21 x 1014 – 2.472 x 1014
= 4.738 x 1014 Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 1014 Hz.
Q 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.
Answer: Wavelength of light produced by the argon laser, λ = 488 nm = 488 × 10−9 m
Stopping potential of the photoelectrons, V0 = 0.38 V
1eV = 1.6 × 10−19 J
∴ V0 = 0.38/1.6 x 10-19 eV
Planck’s constant, h = 6.6 × 10−34 Js
Charge on an electron, e = 1.6 × 10−19 C
Speed of light, c = 3 × 108 m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ0 of the material of the emitter as: eV0 = hc/λ – ø0
∴ ø0 = hc/λ – eV0
= (6.6 x 10-34 x 3 x 108 / 1.6 x 10-19 x 488 x 10-9) – (1.6 x 10-19 x 0.38 / 1.6 x 10-19) = 2.54 – 0.38 = 2.16 eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
Question 11.12: Calculate the
(a) momentum, and
(b) the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer: Potential difference, V = 56 V
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
1/2 mv2 = eV
v2 = 2eV / m
The momentum of each accelerated electron is given as:
p = mv
= 9.1 × 10−31 × 4.44 × 106
= 4.04 × 10−24 kg m s−1
Therefore, the momentum of each electron is 4.04 × 10−24 kg m s−1 .
(b) De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
Q 13. What is the:
(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with a kinetic energy of 120 eV.
Answer: Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) For the electron, we can write the relation for kinetic energy as: Ek = 1/2 mv2
Where, v = Speed of the electron
Momentum of the electron, p = mv = 9.1 × 10−31 × 6.496 × 106 = 5.91 × 10−24 kg ms−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1 .
(b) Speed of the electron, v = 6.496 × 106 m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as: λ = h/p = 6.6 x 10-34 / 5.91 x 10-24 = 1.116 x 10-10 m = 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
Q 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.
Answer: Wavelength of light of a sodium line, λ = 589 nm = 589 × 10−9 m
Mass of an electron, me= 9.1 × 10−31 kg
Mass of a neutron, mn= 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34 Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
K = 1/2 mev2 ….. (1)
We have the relation for de Broglie wavelength as:
Substituting equation (2) in equation (1), we get the relation:
Hence, the kinetic energy of the electron is 6.9 × 10−25 J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
Hence, the kinetic energy of the neutron is 3.78 × 10−28 J or 2.36 neV.
Q 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 × 10-9 kg drifting with a speed of 2.2 m/s?
Answer: (a) Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 × 10−34 Js
De Broglie wavelength of the bullet is given by the relation: λ = h/mv
= 6.6 x 10-34 / 0.040 x 1000 = 1.65 x 10-35 m
(b) Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation: λ = h/mv
= 6.6 x 10-34 / 0.060 x 1 = 1.1 x 10-32 m
(c) Mass of the dust particle, m = 1 × 10−9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation: λ = h/mv
= 6.6 x 10-34 / 2.2 x 1 x 10-9 = 3.0 x 10-25 m
Q 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 the electron.
Answer: Wavelength of an electron (λe) and a photon (λp), λe = λp = λ = 1 nm = 1 × 10−9 m
Planck’s constant, h = 6.63 × 10−34 Js
(a) The momentum of an elementary particle is given by de Broglie relation: λ = h/p
∴ p = h/λ
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.
∴ p = 6.63 x 10-34 / 1 x 10-9 = 6.63 x 10-25 kg m s-1
(b) The energy of a photon is given by the relation: E = hc/λ
Where,
Speed of light, c = 3 × 108 m/s
∴ E = 6.63 x 10-34 x 3 x 108 / 1 x 10-9 x 1.6 x 10-19 = 1243.1 eV
Therefore, the energy of the photon is 1.243 keV.
(c) The kinetic energy (K) of an electron having momentum p, is given by the relation: K = 1/2 p2 / m
Where,
m = Mass of the electron = 9.1 × 10−31 kg
p = 6.63 × 10−25 kg m s−1
∴ K = 1/2 x (6.63 x 10-25)2 / 9.1 x 10-31 = 2.415 x 10-19 J
= 2.415 x 10-19 / 1.6 x 10-19 = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
Q 17. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 x 10-10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.
Answer: (a) De Broglie wavelength of the neutron, λ = 1.40 × 10 −10 m
Mass of a neutron, m n = 1.66 × 10 −27 kg
Planck’s constant, h = 6.6 × 10 −34 Js
Kinetic energy (K) and velocity (v) are related as:
K = 1/2 mnv2 … (1)
De Broglie wavelength (λ) and velocity (v) are related as:
λ = h/ mnv ….(2)
Using equation (2) in equation (1), we get:
Hence, the kinetic energy of the neutron is 6.75 × 10 −21 J or 4.219 × 10 −2 eV.
(b) Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 × 10 −23 kg m 2 s −2 K −1
Average kinetic energy of the neutron:
K’ = 3/2 kT
= 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Q 18. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer: The momentum of a photon having energy (hν) is given as:
p = hv/c = h/λ
λ = h/p …(i)
Where,
λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as: λ = h/mv
But p = mv
Therefore, λ = h/p …(ii)
Where, m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Question 11.19: 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)
Answer:
Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u
But 1 u = 1.66 × 10 −27 kg
Therefore, m = 28.0152 ×1.66 × 10 −27 kg
Planck’s constant, h = 6.63 × 10 −34 Js
Boltzmann constant, k = 1.38 × 10 −23 J K −1
We have the expression that relates mean kinetic energy (3/2 kT ) of the nitrogen molecule with the root mean square speed (vrms) as:
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
Question 11.20: (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 e/m is given to be 1.76 × 1011 C kg–1.
(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?
Answer: (a) Potential difference of the evacuated tube = 500 V
Specific charge of the electron, e/m = 1.76 × 1011 C kg–1
Kinetic energy = (1/2) mv2 = eV
Speed of the emitted electron, v = (2Ve/m)1/2
= (2 x 500 x 1.76 x 1011)1/2
= 1.32 x 107 m/s
(b) Collector potential, V = 10 MV = 10 x 106 V.
Speed of electron = v = (2Ve/m)1/2
= (2 x 107 x 1.76 x 1011)1/2
= 1.88 x 109 m/s
This answer is not correct. Since the value is greater than the speed of light (c). The expression (1/2) mv2 for energy should be used in the non -relativistic limit.i.e., v << c.
In the relativistic limits, the total energy is given as
E = mc2
Here,
m is the relativistic mass
m = m0 (1- v2/c2)1/2
m0 = mass of the particle at rest
Kinetic energy is given as
K = mc2 – m0c2
Q 21. (a) A monoenergetic electron beam with an electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 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 × 1011C kg–1.
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Answer: (a) Given: The speed of electron of monoenergetic electron beam is 5.20× 10 6 ms -1 , the magnetic field is 1.30× 10 −4 T normal to the beam velocity.
The force exerted on the electron is given as,
F=evBsinθ(1)
Where, the velocity of the electron is v, magnetic field experienced by the electron is B, angle between velocity of electron and direction of magnetic field is θ, charge on the electron is e , and mass of the electron is m.
The magnetic force provides centripetal force to the beam due to which it traces a circular path.
F c , the centripetal force and r, the radius of circular path is given by,
F c = m v 2 r (2)
For equilibrium, equation (1) is equal to (2),
evBsinθ= m v 2 r r= v ( e m )Bsinθ (3)
The value of e/m is 1.76× 10 11 Ckg −1 .
By substituting the given values in the above expression, we get,
r= 5.20× 10 6 ( 1.76× 10 11 )×1.30× 10 −4 ×sin 90 0 =0.227 m
Thus, the radius of the circle traced by beam is 0.227 m.
(b) Given: The energy of the electron beam is 20 MeV.
The energy of the electron is given as,
E= m v 2 2 v= ( 2E m ) 1 2
The value of m is 9.1× 10 −34 kg.
By substituting the given values in the above expression, we get,
v= ( 2×20×1.6× 10 −13 9.1× 10 −34 ) 1 2 =2.652× 10 9 ms -1
The velocity of electron is greater than speed of light ( 3× 10 8 ms -1 ), which is not possible. So, this formula can’t be used. When very high speed is considered, the relativistic domain comes into consideration.
The mass of the particle is given as,
m= m 0 [ 1− v 2 c 2 ] 1 2 (4)
Where, the mass of the particle at rest is m 0 and speed of light is c.
By substituting the given values in the above expression, we get,
r= m 0 vc eBsinθ [ c 2 − v 2 ] 1 2
Thus, the modified formula for radius of circular path is r= m 0 vc eBsinθ [ c 2 − v 2 ] 1 2 .
Q 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 (∼10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit 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 ‘fine beam tube’ method.) Determine e/m from the data.
Answer: It is given that the potential of an anode, V=100 V, the magnetic field experienced by the electrons, B=2⋅83× 10 −4 Tand the radius of the circular orbit, r=12⋅0 cm.
The energy of each electron is equal to its kinetic energy.
1 2 m v 2 =eV v 2 = 2eV m (1)
From the formula of centripetal force,
F= m ν 2 r
The centripetal force will be equal to magnetic force and it can be written as,
m v 2 r =evΒ v= eΒr m (2)
Substitute the value of vin equation (1).
2eV m = ( eΒr m ) 2 e m = 2V Β 2 r 2
Substitute the values in above expression.
e m = 2×100 ( 2⋅83× 10 −4 ) 2 × ( 12× 10 −2 ) 2 =1⋅73× 10 13 Ckg −1
Hence, the specific charge ratio ( e m ) is 1⋅73× 10 13 Ckg −1 .
Question 11. 23: (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. 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?
Answer: (a) Wavelength produced by an X-ray tube, 
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
The maximum energy of a photon is given as:
Therefore, the maximum energy of an X-ray photon is 27.6 keV.
(b) Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.
Question 11.24:In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)
Answer: It is given that the total energy of two y-rays is 10⋅2 BeV or 10⋅2× 10 9 ×1⋅6× 10 −19 J.
The energy for each γ-ray is,
E’= E 2 E’=8⋅16× 10 −19 J
The formula of maximum energy of each γ-ray is,
E’= hc λ
Substitute the values in above expression.
8⋅16× 10 −19 = 6⋅626× 10 −34 ×3× 10 8 λ λ=2⋅436× 10 −16 m
Hence, the value of wavelength associated with each γ-ray is 2⋅436× 10 −16 m.
Question 11.25: 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 ‘count photons’, 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 (∼10–10 W m–2). Take the area of the pupil
to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz
Answer:
(a) Power of the medium wave transmitter, P = 10 kW = 104 W
Energy emitted by the transmitter per second, E = 104
Wavelength of the radio waves, λ = 500 m
The energy of the wave is given as, E’ = hc/λ
E’ = (6.6 x 10-34 x 3 x 108)/500
= 3.96 x 10-28 J
Let n be the number of photons emitted by the transmitter
nE’ = E
n = E/E’
= 104/(3.96 x 10-28 )
= 0.2525 x 1032
The energy E’ of the radio photon is very less, but the number of photons emitted is large. The total energy of the radio ways can be considered as continuous and the existence of the minimum quantum energy can be ignored.
(b) Intensity of the light perceived by the human eye, I = 10–10 W m–2
Area of the pupil, A = 0.4 cm2 = 0.4 x 10-4 m2
Frequency of the white light, ν = 6 x 1014 Hz
h = Planck’s constant = 6.6 x 10-34 Js
Energy of the emitted photon, E = hν
= 6.6 x 10-34 x6 x 1014
= 3.96 x 10-19 J
Let n be the total number of photons falling per unit area per unit time. The total energy per unit for n photons is
E = n x 3.96 x 10-19 J/s/m2
Total energy per unit for n photons is equal to the intensity of the light.
E = I
I = n x 3.96 x 10-19 J/s/m2
n = I/3.96 x 10-19
= 10-10/3.96 x 10-19
= 2.52 x 108 m2/s
The total number of photons entering the pupil is given as,
nA = 2.52 x 108 x 0.4 x 10-4
= 1.008 x 104 s-1
The number is large. So the human eye can never count the number of individual photons.
Q 26. Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer: Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m
Stopping potential of the metal, V0 = 1.3 V
Planck’s constant, h = 6.6 × 10−34 J
Charge on an electron, e = 1.6 × 10−19 C
Work function of the metal = ∅o
Frequency of light = ν
We have the photo-energy relation from the photoelectric effect as:
∅o = hν − eV0
Let ν0 be the threshold frequency of the metal.
∴ ∅o = hν0
Wavelength of red light, λ=6328 A = 6328 × 10−10 m
∴Frequency of red light,
Since ν0> νr, the photocell will not respond to the red light produced by the laser.
Q 27. Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium 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 photo-cell. Predict the new stopping voltage.
Answer: Wavelength of the monochromatic radiation, λ = 640.2 nm
= 640.2 × 10−9 m
Stopping potential of the neon lamp, V0 = 0.54 V
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.6 × 10−34 Js
Let ∅o be the work function and ν be the frequency of emitted light.
We have the photo-energy relation from the photoelectric effect as:
eV0 = hν − ∅o
Wavelength of the radiation emitted from an iron source, λ‘ = 427.2 nm
= 427.2 × 10−9 m
Let Vo be the new stopping potential. Hence, photo-energy is given as:
Hence, the new stopping potential is 1.50 eV.
Q 28. A mercury lamp is a convenient source for studying the 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:
λ1 = 3650 Å,
λ2 = 4047 Å,
λ3 = 4358 Å,
λ4 = 5461 Å,
λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be: V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V. Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Answer: It is given that the wavelengths, λ 1 =3650 A o , λ 2 =4047 A o , λ 3 =4358 A o , λ 4 =5461 A o
λ 5 =6907 A o , the stopping voltages corresponding to wavelengths, V 01 =1⋅28 V,
V 02 =0⋅95 V, V 03 =0⋅74 V, V 04 =0⋅16 V, V 05 =0 V
The formula of photo energy relation can be written as,
ϕ=hυ−e V 0 V 0 = h e υ− ϕ e (1)
The formula of frequency is,
υ= c λ
Calculate the values of frequencies for given wavelengths respectively.
υ 1 = c λ 1 = 3× 10 8 3650× 10 −10 =8⋅219× 10 14 Hz
υ 2 = c λ 2 = 3× 10 8 4047× 10 −10 =7⋅412× 10 14 Hz
υ 3 = c λ 3 = 3× 10 8 4358× 10 −10 =6⋅884× 10 14 Hz
υ 4 = c λ 4 = 3× 10 8 5461× 10 −10 =5⋅493× 10 14 Hz
υ 5 = c λ 5 = 3× 10 8 6907× 10 −10 =4⋅343× 10 14 Hz
Draw the table as given below.
| Frequency × 10 14 Hz | 8⋅219 | 7⋅412 | 6⋅884 | 5⋅493 | 4⋅343 |
| Stopping potential V 0 | 1⋅28 | 0⋅95 | 0⋅74 | 0⋅16 | 0 |
Draw a graph between stopping potential and frequency as per given values.
The graph intersects the frequency axis at 5× 10 14 Hz, so this frequency will be the threshold frequency.
From the graph, the slope can be calculated as,
AB CB = 1⋅28−0⋅16 ( 8⋅214−5⋅493 )× 10 14
This slope will be equal to slope of equation (1) as h e .
h e = 1⋅28−0⋅16 ( 8⋅214−5⋅493 )× 10 14
Substitute the value of e in the above equation.
h= 1⋅12×1⋅6× 10 −19 ( 8⋅214−5⋅493 )× 10 14 h=6⋅6× 10 −34 Js
The formula of work function is,
ϕ=h υ 0
Substitute the values in above expression.
ϕ=6⋅6× 10 −34 ×5× 10 14 =3⋅286× 10 −19 J = 3⋅286× 10 −19 1⋅6× 10 −19 =2⋅1 eV
Hence, the value of Planck’s constant is 6⋅6× 10 −34 Js, the threshold frequency is
5× 10 14 Hz and the work function is 2⋅1 eV.
Q 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 radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Answer: Wavelength λ = 3300 Å
Speed of light = 3 x 108 m/s
Planck’s constant = 6.63 x 10-34 Js
Energy of the photon of the incident light
E = hc/λ = (6.63 x 10-34 x 3 x 108)/3300x 10-10
⇒ 6.018 x 10-19 J
⇒ (6.018 x 10-19 J)/1.6 x 10-19
= 3.7 eV
The energy of the incident radiation is greater than the work function of Na and K. It is lesser for Mo and Ni. Therefore, Mo and Ni will not show photoelectric effect.
If the laser is brought nearer and placed 50 cm away, then the intensity of the radiation will increase. The energy of the radiation will not be affected. Therefore, the result will be the same. However, the photoelectrons from Na and K will increase in proportion to intensity.
Q 30. Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the 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?
Answer: Intensity of the light = 10–5 W m–2
Surface area of the sodium photocell, A = 2 cm2
Incident power of the light, P = I x A
= 10-5 x 2 x 10-4
= 2 x 10-9 W
Work function of the metal, Φ0 = 2 eV
= 2 x 1.6 x 10-19
= 3.2 x 10-19 J
The number of layers of sodium that absorbs the incident energy, n = 5
Atomic area of the sodium atom, Ae is 10-20 m2
Hence, the number of conduction electrons in n layers is given as:
n’ = n x (A/Ae)
= 5 x [(2 x 10-4)/10-20] = 1017
The incident power is absorbed by all the electrons continuously. The amount of energy absorbed per electron per second is
E = P/n’
= (2 x 10-9)/1017
= 2 x 10-26 J/s
The time for photoelectric emission
t = Φ0/E
= (3.2 x 10-19)/(2 x 10-26) = 1.6 x 107 s ≈ 0.507 years
The time required for the photoelectric emission is almost half a year. This is not practical. Therefore, the wave function is in disagreement with the given experiment.
Q 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 Å, which is of the order of interatomic spacing in the lattice) (me =9.11 × 10–31 kg).
Answer: An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Å = 10−10 m
Mass of an electron, me = 9.11 × 10−31 kg
Planck’s constant, h = 6.6 × 10−34 Js
Charge on an electron, e = 1.6 × 10−19 C
The kinetic energy of the electron is given as:
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
Energy of a photon,
Hence, a photon has a greater energy than an electron for the same wavelength.
Q 32. (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in question 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 × 10–27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer: (a) Given: The kinetic energy of the neutron is 150 eV and the mass of the neutron is 1.675× 10 −27 kg.
We know that,
1 eV=1.6× 10 −19 J
The kinetic energy of the neutron is given as,
K= 1 2 m n v 2
Where, m n is the mass of neutron and v is the velocity of the neutron.
Re-arranging the above equation, we get
m n v= 2K m n
de Broglie wavelength is given as,
λ= h p = h m n v = h 2K m n
Where, h is the Planck’s constant, p is the momentum and λ is the de Broglie wavelength.
By substituting the given values in the above equation, we get
λ= 6.626× 10 −34 2×150 ×1.6× 10 −19 ×1.675× 10 −27 =2.337× 10 −12 m
Thus, the de Broglie wavelength of neutron is 2.337× 10 −12 m.
In the previous problem, it is given that the inter-atomic spacing is about 1 A 0 or 10 −10 m. This shows that inter atomic separation is 100 times more than the wavelength.
Thus, a neutron beam of energy 150 eV is not suitable for diffraction experiments because for diffraction experiment, inter atomic separation should be of same order as the wavelength.
(b) Given: The temperature of neutron beam is 27 °C.
The average kinetic energy of neutron is given as,
K= 3 2 kT
Where, k is the Boltzmann’s constant and T is the given temperature.
The de Broglie wavelength is given as,
λ= h 2 m n K = h 3 m n kT
By substituting the given values in the above equation, we get
λ= 6.626× 10 −34 3×1.675× 10 −27 ×1.38× 10 −23 ×( 273+27 ) =1.45× 10 −10 m
Since, this De Broglie wavelength is comparable to the inter-atomic spacing of a crystal. Thus, the high energy neutron beam should be thermalised before it can be used for neutron diffraction experiments.
Q 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?
Answer: Given: The electrons are accelerated by a voltage, 50 kV.
The kinetic energy of the electron is given as,
E=eV
Where, e is the charge of electron and V is the voltage.
By substituting the given values in the above equation, we get
E=1.6× 10 −19 ×50× 10 3 =8× 10 −15 J
The de Broglie wavelength is given as,
λ= h 2 m e E
By substituting the value in the above expression, we get
λ= 6.6× 10 −34 2×9.1× 10 −31 ×8× 10 −15 =5.467× 10 −12 m
Wavelength of yellow light is 5.9× 10 −7 m.
Thus, wavelength of an electron is nearly 10 5 times less than the wavelength of yellow light.
Since, the resolving power of a microscope is inversely proportional to the wavelength of light used, the resolving power of an electron microscope will be 10 5 times that of an optical microscope.
Q 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 10–15 m or less. This structure was first probed in the early 1970s 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.)
Answer: Wavelength of the proton or neutron, λ ≈ 10-15 m
Rest mass-energy of an electron :
m0c2 = 0.511 MeV
= 0.511 x 106 x 1.6 x 10-19
= 0.8176 x 10-13 J
Planck’s constant, h = 6.6 x 10-34 Js
Speed of light, c = 3 x 108 m/s
The momentum of the proton or a neutron is given as
p = h/λ
= 6.6 x 10-34 /10-15
= 6.6 x 10-19 kg m/s
The relativistic relation for energy (E) is given as
E2 = p2c2 + m20C4
= (6.6 x 10-19 x 3 x 108)2 + (0.8176 x 10-13)2
= 392.04 x 10-22 + 0.6685 x 10-26
≈ 392.04 x 10-22
⇒ E = 19.8 x 10-11
= 19.8 x 10-11/1.6 x 10-19
= 12.375 x 108 eV
Order of energy of these electron beams is 12.375 x 108 eV
Q 35. Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
Answer: Given: The temperature is 27 °C and atmospheric pressure is 1 atm.
The mass of helium atom is given as,
m= Atomic mass N A
Where, N A is the Avogadro’s number.
By substituting the values in the above equation, we get
m= 4 6.023× 10 23 =6.64× 10 −24 g =6.64× 10 −27 kg
The average kinetic energy of helium atom is given as,
E= 3 2 kT
Where, k is the Boltzmann’s constant and T is the temperature.
The de Broglie wavelength is given as,
λ= h p = h mv = h 2mE = h 3mkT
Where, h is the Planck’s constant and p is the momentum and λ is the de Broglie wavelength.
By substituting the values in the above equation, we get
λ= 6.626× 10 −34 3×6.64× 10 −27 ×1.38× 10 −23 ×( 273+27 ) =0.729× 10 −10 m
Thus, the typical de Broglie wavelength of electron is 0.729× 10 −10 m.
From the ideal gas law,
PV=Nkt V N = kT P
Where, V is the volume and N is the number of moles of the gas.
The above equation can be re-written as,
V N = kT P
The mean separation between two atoms is given as,
r= ( V N ) 1 3 = ( kT P ) 1 3
By substituting the values in the above equation, we get
r= ( 1.38× 10 −23 ×300 1.0135× 10 5 ) 1 3 =3.44× 10 −9 m
Thus, mean separation between two atoms of the gas is greater than the de Broglie wavelength of the gas.
Q 36. Compute the typical de Broglie wavelength of an electron in metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishability has many fundamental implications which you will explore in more advanced Physics courses.]
Answer: Given: The mean separation between two electrons in a metal is 2× 10 −10 m and the temperature is 27 °C
de Broglie wavelength of an electron is given as,
λ= h 3mkT (1)
Where, Planck’s constant is h, the mass of electron is m, Boltzmann Constant is k, the temperature is T and the wavelength is λ.
By substituting the given values in the above formula, we get
λ= 6.6× 10 −34 3×9.1× 10 −31 ×1.38× 10 −23 ×300 = 6.6× 10 −34 11302.2× 10 −54 = 6.6× 10 −34 106.311× 10 −27 =6.2× 10 −9 m
Thus, the de Broglie wavelength of electron is 6.2× 10 −9 m which is much greater than the given inter-electron separation.
Q 37. Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m 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:
E = h ν, p = h/λ
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?
Answer: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. This is because the nuclear forces grow stronger if they are pulled apart. Therefore, it seems that fractional charges may exist in nature. The observable charges are an integral multiple of electrical charges (e).
(b) The relation between electric field and the magnetic field,
eV = (1/2) mv2 and eBv = mv2/r
Here,
e = electric charge
v = velocity
V = potential
r = Radius
B = magnetic field
From these equations, it can be understood that the dynamics of an electron can be determined only by the ratio e/m and not by e and m separately.
(c) At the atmospheric pressure, the ions in the gas does not have a chance of reaching their respective electrons due to collision and recombination with other molecules in the gas. At low pressures, ions have a chance to reach their respective electrons and results in the flow of current.
(d) The minimum energy required for an electron in the conduction band to get out of the metal is called work function. These electrons occupy different energy levels, because of which for the same incident radiation, electrons come out with different energies.
(e) The absolute value of the energy of a particle is arbitrary within the addictive constant. Therefore, the wavelength(λ) is significant, but the frequency (ν) of the electron does not have direct physical significance. Therefore, product νλ has no physical significance.