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NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter
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Dual Nature of Radiation and Matter
Chapter: 11
| Part – II |
EXERCISE
1. Find the:
(a) maximum frequency. and
Ans: Given:
V = 30 KV = 3 × 103 V
e = 1.6 × 10-19 C
E = eV = hu
v = eV/h
= (1.6 × 10-19 × 30 × 103) / (6.63 × 10-34)
= 7.2 × 1018 Hz.
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Ans: Given:
λ = c/v = 3 × 108 / 7.2 × 1018
= 0.414 × 10-10
= 0.0414 × 10-9 m
= 0.0414 nm.
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.
Ans: Given:
W = 2.14 eV
= 2.14 × 1.6 × 10-19 J.
= 3.424 × 10-19 J
v = 6 × 1014 Hz
h = 6.626 × 10-34 Js.
K.E. = hv – W
= (6.626 × 10-34 Js) × (6 × 1014 Hz) – 3.424 × 10-19 J.
= (3.976 – 3.424) × 10-19
= 0.55 × 10-19 J
= 0.55 × 10-19 / 1.6 × 10-19 eV
= 0.344 eV.
(b) Stopping potential. and
Ans: K = eV
= Vo = K/e
= (0.345 × 1.6 × 10-19) / (1.6 × 10-19)
= 0.343 V.
(c) Maximum speed of the emitted photoelectrons?
Ans: Given:
KEmax = (1/2) mv2 (Here m = 9.11 × 10-31 kg)
v2 = 2KEmax / m
v = √(2× 3.45 × 10-20 J / 9.11 × 10-31 kg )
= 8.69 × 105 m/s
3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Ans: Given:
e = 1.602 × 10-19 c
Vstop = 1.5 V
Kmax = 1.602 × 10-19 C × 1.5 V
= 2.403 × 10-19 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.
Ans: λ: 632.8 nm = 632.8 × 10-9 m
P: 9.42 mW = 9.42 × 10-3 W
c = 3 × 108 m/s
h = 6.626 × 10-34 J-s
mH = 1.67 × 10-27 kg
E = hc /λ
= (6.626 × 10-34 J) × (3 × 108 m/s) / 632.8 × 10-9m
= 3.14 × 10-19 J.
P = h/λ = 6.626 × 10-34 / 632.8 × 10-9
= 1.05 × 10-27 kg ms-1
(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
Ans: Given:
N = P/E
= (9.42 × 10-3) × (3.15 × 10-19 m/s)
= 2.99 × 1016.
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Ans: Given:
mv = P
v = P/m
= 1.05 × 10-27 / 1.67 × 10-27
= 0.628 ms-1.
5. 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: Given:
(e = 1.602 × 10-19 C)
Ek = eV0
eV0 = hf – ø
V0 = hf – ø
Slope = h/e
(Given that the slope 4.12 × 10-15)
h = slope × e
= (4.12 × 10-15 V s) × (1.602 × 10-19 C)
= 6.60 × 10-34 J s.
6. 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: Given:
𝑓 = 8.2 × 1014 Hz
h = 6.626 × 10-34 J s
e = 1.602 × 10-19 C
ϕ = h𝑓0
= 6.626 × 10-34 J s ) × (3.3 × 1014 Hz)
= 2.19 × 10-19 J.
Ek = E – ø
= (5.44 × 10-19 J) – (2.19 × 10-19 J)
= 3.25 × 10-19 J.
Ek = eV0
V0 = Ek/e
= 3.25 × 10-19 J / (1.602 × 10-19 C)
= 2.03 V.
7. 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: Given:
h = 4.1357 × 10-15 eV
c = 3 × 108 m/s
λ = 330 nm = 330 × 10-9 m
E = hc / λ
= (4.1357 × 10-15 eV) × (3 × 108 m/s) / 330 × 10-9 m
= 3.76 eV.
8. 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: Given:
m = 9.11 × 10-31 kg
vmax = 6.0 × 105 m/s
KE = 1/2 mv2max
KE = (1/2 × 9.11 × 10-31 kg) × (6.0 × 105 m/s)2
= 1.64 × 10-19 J.
Convert this energy into electron volts (eV) electron ( 1.602 × 10-19 C)
KE = (1.64 × 10-19 J) / (1.602 × 10-19 C)
= 1.02 eV.
Energy of the incident photons:
𝑓 = 7.21 × 1014 Hz
h = 4.1357 × 10-15 eV
Ephoton = hf
= (7.21 × 1014 Hz) × (4.1357 × 10-15)
= 2.98 eV.
Determine the threshold frequency:
ø = Ephoton – K
= 2.98 eV – 1.02 eV
= 1.96 eV
fthreshold = Ø/h
= 1.96 eV / 4.1357 × 10-15 eV
= 4.74 × 1014 Hz.
9. 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: Given:
λ = 488 nm = 488 × 10-9 m
V0 = 0.38 V
h = 6.626 × 10-34 Js
e = 1.6 × 10-19 C
C = 3 × 108 ms-1
V = C/λ
= (3 × 108) / 488 × 10-9
= 6.15 × 1014 Hz
eV0 = hv – W
= hv – eV
= 6.626 × 10-34 × 6.15 × 1014 – 1.6 × 10-19 × 0.38
= 3.467 × 10-19 J
= 3.467 × 10-19 / (1.6 × 10-19)
= 2.167 eV.
10. What is the de Broglie wavelength of:
(a) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.
Ans: Here:
P = mv
= 0.040 kg × 1000 m/s = 40 kg
λ = h/p
= (6.626 × 10-34 m2 kg/s) / ( 40 kg )
= 1.66 × 10-35 m.
(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and.
Ans: Given:
p = mv = 0.060 kg × 1.0 m/s
= 0.060 kg
λ = h/p
= (6.626 × 10-34 m2 kg/s) / (0.060 kg)
= 1.10 × 10-32 m.
(c) A dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s?
Ans: Given:
p = mv = 1.0 × 10-9 kg × 2.2 m/s
= 2.2 × 10-9 kg
λ = h/p
= (6.626 × 10-34 m2 kg/s) / (2.2 × 10 -9 kg)
= 3.01 × 10-25 m
11. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Ans: Given:
P = hv/c = h/λ
λ = h/p
de-broglie wavelength of photon:
λ = h/mv = h/p
= h / (hv / c)
= c/v.
Thus, the wavelength of electromagnetic radiation is equal to the de-broglie wavelength.
