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Asked by Nishit Last Modified
Answer (i) Energy (E) of a photon = hν
Where,
h = Planck’s constant = 6.626 × 10–34 Js
c = velocity of light in vacuum = 3 × 108 m/s
λ = wavelength of photon = 4 × 10–7 m
Substituting the values in the given expression of E:
Hence, the energy of the photon is 4.97 × 10–19 J.
(ii) The kinetic energy of emission Ek is given by
= (3.1020 – 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photoelectron (ν) can be calculated by the expression,
Where, 
is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:
v = 5.84 × 105 ms–1
Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.
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