Question

The cutoff frequency for a certain element is 1.22 x 1015 Hz. What is its work function in eV?
Hint: 1 eV 1.60 x 10-19 J
=[?] eV

someone please teach me how to solve this its gonna make me cry

Answer 1

The work function of the metal is obtained as 5.1 eV

What is the cutoff frequency?

The cutoff frequency is the frequency below which photo electric effect can not occur as electrons are not removed from the metal surface.

Now;

fo = 1.22 x 10^15 Hz

Wo = hfo

Wo = 6.6 * 10^-34 * 1.22 x 10^15

Wo = 8.1 * 10^-19 J

If 1 eV = 1.60 x 10-19 J

x eV = 8.1 * 10^-19 J

x = 8.1 * 10^-19/ 1.60 x 10-19

x = 5.1 eV

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