Question

When a light of frequency 9 × 10¹⁴ Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8 × 10⁵ ms⁻¹. Determine the threshold frequency of the surface.

Answer 1

v = 9 × 10¹⁴ Hz

v = 8 × 10⁵ ms⁻¹

From Einstein’s photo electric equation

hv = hv₀ – `1/2` mv²

hv₀ = hv – `1/2` mv²

v₀ = `("hv" - 1/2 "mv"^2)/"h"`

v₀ = `"v" - ("mv"^2)/(2"h")`

= `9 × 10^14 - (9.1 xx 10^-31 xx 64 xx 10^10)/(2 xx 6.626 xx 10^-34)`

= `9 xx 10^14 - (582.4 xx 10^-21)/(13.252 xx 10^-34)`

= 9 × 10¹⁴ − 4.39 × 10¹⁴

= (9 − 4.39) × 10¹⁴

v₀ = 4.61 × 10¹⁴ Hz