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Q1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons. Ans: 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.
Q2 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, and (c) maximum speed of the emitted photoelectrons? Ans: 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.
Q3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? Ans: Photoelectric cut-off voltage, V0 = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as: Ke = eVo
Where,
e = Charge on an electron = 1.6 × 10−19 C
∴ Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10−19 J.
Q4 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 in order to have the same momentum as that of the photon? Ans: 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
Q5 The energy flux of sunlight reaching the surface of the earth is 1.388 × 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. Ans: 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.
Q6 In an experiment on 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. Ans: 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 Vs
V 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
Q7 A 100 W 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? Ans: Power of the sodium lamp, P = 100 W
Wavelength 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.
Q8 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2×1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission. Ans: 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
Q9 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? Ans: 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.
Q10 Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons? Ans: 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.
Q11 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. Ans: 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.
Q12 Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. Ans: 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.
Q13 What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV. Ans: 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.
Q14 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. Ans: 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.
Q15 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? Ans: (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
Q16 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. Ans: 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.
Q17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 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. Ans: (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.
Q18 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). Ans: 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.
Q19 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) Ans: 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.
Class 12 Physics Chapter 11: Dual Nature Of Radiation And Matter - NCERT Solutions
This page focuses on the complete NCERT solutions for Class 12 Physics Chapter 11: Dual Nature Of Radiation And Matter. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Dual Nature Of Radiation And Matter question answers will offer you valuable insights and explanations.
The eleventh chapter in the Class 12 Physics NCERT book, you study about the nature of radiation and the properties. This chapter also teaches you about the photoelectric effects of light, dual nature of radiation, photoelectric equation of Einstein, the Wave Theory, Wave Nature of matters around us, Energy quantum of radiation, etc. Saralstudy provides you with the detailed reference solutions ready for you for understanding the chapters completely.
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(a) Estimate the number of electrons transferred (from which to which?)
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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
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A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
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A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
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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, and
(c) maximum speed of the emitted photoelectrons?
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