Question

The threshold wavelength of a metal is 450 nm. Calculate

1. the work function of the metal in eV and
2. the maximum energy of the ejected photoelectrons in eV by incident radiation of 250 nm.

Answer 1

Given: Threshold wavelength (λ₀) = 450 × 10⁻⁹ m

Incident wavelength (λ) = 250 × 10⁻⁹ m

i. Work function (Φ) = `(hc)/(lambda_0)`

= `(6.63 xx 10^-34 xx 3 xx 10^8)/(450 xx 10^-9)`

= 4.42 × 10⁻¹⁹ J

= `(4.42 xx 10^-19)/(1.6 xx 10^-19)`

= 2.76 eV

ii. Maximum kinetic energy (KE_max) = `(hc)/lambda - Phi`

= `(6.63 xx 10^-34 xx 3 xx 10^8)/(250 xx 10^-9) - 4.42 xx 10^-19`

= 7.96 × 10⁻¹⁹ − 4.42 × 10⁻¹⁹

= 3.54 × 10⁻¹⁹ J

= `(3.54 xx 10^-19)/(1.6 xx 10^-19)` eV

= 2.21 eV