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 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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"`
