Question

The work function of the metal is 2.13 eV. A photon having wavelength of 4 × 10^-7m strikes the metal then calculate kinetic energy of emission of energy of photon and velocity of photoelectron.

Answer 1

(i) Energy (E) of a photon = hv
= hc/λ
Where, h = Planck’s constant = 6.626 × 10^-34 J sec
c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = wave length of photon = 4 × 10^-7 m
Substituting the values, we get,
E = (6.626 × 10³⁴ )(3 × 10⁸)/(4 × 10^-7)
= 4.965 × 10^-19 J
Hence, the energy of the photon is 4.97 × 10^-19 J.

(ii) Kinetic energy of the emission = hv – hv_o
= (E – W)
= (4.97 × 10^-19 /1.6020 × 10^-19) eV – 2.13 eV
= (3.1020 – 2.13)eV
= 0.97 eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

where (hv – hv_o) is kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.
Substituting the values in the given expression of v:

Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^-1