Question

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, v0. Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 3.09 Times 10^14 s ^-1. With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of lambda = 225 nm? KEelectron = .

Answer 1

frequency f = 3.09 X 10¹⁴ /sec

h = 6.67 X 10^-34 J.s

a )

the minimum energy is E

E = h f

E = 6.67 X 10^-34 X 3.09 X 10¹⁴

E = 20.163 X 10^-20 J

E = 2.0163 X 10^-19 J or

E = 2.0163 X 10^-19 / 1.6 X 10^-19

E = 1.26 eV

b )

$\lambda$ = 225 nm

$\lambda$ = 225 X 10^-9 m

we have f = c / $\lambda$

f = 3 X 10⁸ / 225 X 10^-9

f = 1.33 X 10¹⁵ /sec

E = h f

E = 6.67 X 10^-34 X 1.33 X 10¹⁵

E = 8.89 X 10^-19 J

or

E = 8.89 X 10^-19 / 1.6 X 10^-19

E = 5.55 eV