Question

UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength is 4965 Å. Determine the maximum energy of the electron emitted.

Answer 1

λ = 1800 × 10⁻¹⁰ m

λ₀ = 4965 × 10⁻¹⁰m

h = 6.6 × 10⁻³⁴ Js

c = 3 × 10⁸ ms⁻¹

Maximum kinetic energy of electron,

K_max = `"hc" (1/λ - 1/λ_0)`

= `(6.6 xx 10^-34 xx 3 xx 10^8)/(10^-10) xx (1/1800 - 1/4965)`

= `19.8 xx 10^-16 xx 3165/(8937 xx 10^3)`

= 7.01208 × 10⁻¹⁹ J

= `(7.01208 xx 10^-19)/(1.6 xx 10^-19)`

= 4.38 eV

K_max = 4.40 eV