Question

A photon of wavelength 4 × 10^–7 m strikes on the metal surface, the work function of the metal being 2.13 eV. Calculate

1. the energy of the photon (eV),
2. the kinetic energy of the emission, and
3. the velocity of the photoelectron (1 eV = 1.6020 × 10^–19 J).

Answer 1

(i) Energy (E) of a photon = `"hν" = "hc"/lambda`

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:

`"E" = ((6.626xx10^(-34))(3xx10^8))/(4xx10)^(-7)` = `4.9695 xx 10^(-19)  "J"`

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission E_k is given by

`= "hv" - "hv"_0`

`=("E" - "W")"eV"`

`= ((4.9695xx10^(-19))/(1.6020xx10^(-19))) "eV" - 2.13 " eV"`

= (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,

`1/2 "mv"^2 = "hv" - "hv"_0`

`=> "v" = sqrt((2("hv" - "hv"_0))/"m")`

where (`"hv" - "hv"_0`) 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" = sqrt((2xx(0.9720xx1.6020xx10^(-19))"J")/(9.10939xx10^(-31)"kg"`

`= sqrt(0.3418 xx 10^12 "m"^2"s"^(-2))`

v = 5.84 × 10⁵ ms^–1

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