When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.

(Use h = 6.63 × 10^{-34}J-s = 4.14 × 10^{-15} eV-s, c = 3 × 10^{8} m/s and me = 9.1 × 10^{-31}kg)

#### Solution

Given :-

Wavelength of light, `λ = 400 "nm" = 400 xx 10^-9 "m"`

Stopping potential, `V_0 = 1.1 V`

From Einstein's photoelectric equation,

`(hc)/λ = (hc)/(λ_0) + eV_0`,

where h = Planck's constant

c = speed of light

λ = wavelength of light

`λ_0` = threshold wavelength

`V_0` = stopping potential

On substituting the respective values in the above formula , we get :

`(6.63 xx 10^-34 xx 3 xx 10^8)/(400 xx 10^-9) = (6.63 xx 10^-34 xx 3 xx 10^8)/λ_0 + 1.6 xx 10^-19 xx 1.1`

`⇒ 4.97 xx 10^-19 = (19.89 xx 10^-26)/λ_0 + 1.76 xx 10^-19`

`⇒ 4.97 = (19.89 xx 10^-7)/λ_0 + 1.76`

`⇒ (19.89 xx 10^-7)/λ_0 = 4.97 - 1.76 = 3.21`

`⇒ λ_0 = (19.89 xx 10^-7)/3.21`

`= 6.196 xx 10^-7 "m" = 620 "nm"`