Question

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Work function of a photoelectric material, ϕ = 4 eV = 4 × 1.6 × 10⁻¹⁹ J

Stopping potential, V₀ = 2.5 V

Planck's constant, `h = 6.63 xx 10^-34  "Js"`

(a) Work function of a photoelectric material,

`phi = (hc)/λ_0`

Here, λ₀ = threshold wavelength of light

c = speed of light

`therefore λ_0 = (hc)/phi`

`λ_0 = (6.63 xx 10^-34 xx 3 xx 10^8)/(4 xx 1.6 xx 10^-19)`

`λ_0 = (6.63 xx 3)/64 xx (10^27)/(10^-9)`

`λ_0 = 3.1 xx 10^-7  "m"`

`λ_0 = 310  "nm"`

(b) From Einstein's photoelectric equation,

`E = phi + eV_0`

On substituting the respective values , we get :-

`(hc)/λ = 4 xx 1.6 xx 10^-19 + 1.6 xx 10^-19 xx 2.5`

`⇒ λ = (6.63 xx 10^-34 xx 3 xx 10^8)/(6.5 xx 1.6 xx 10^-19)`

`⇒ λ = (6.63 xx 3 xx 10^-26)/(1.6 xx 10^-19 xx 6.5)`

`⇒ λ = 1.9125 xx 10^-7 = 191  "nm"`