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.
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.
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.
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.
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
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.
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
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.
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
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.
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.
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.
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.
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.
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.
(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
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.
(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.
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.
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.