Mamta k.

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(1)We know λ = 4 × 10^–7 m(given)

C = 3 x 108

From the equation E= hv or hc/ λ

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, (hv-hv₀) 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.