Question

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV .

What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 330 nm falls on the same surface?

Use h = 6.63×10^?34J?s for Planck's constant and c = 3.00×10⁸m/s for the speed of light and express your answer in electron volts.

Answer 1

maximum kinetic energy of an ejected electron

Kmax = hc/1 - W

so work function

W = hc/1 - Kmax

W = [ 6.63*10^-34 *3*10⁸ / 400*10^-9 ] - [1.10*1.6*10^-19 ]

W = 4.97*10^-19 - 1.76*10^-19

W = 3.21*10^-19 J

Now

Maximum kinectic energy of an ejected electron

Kmax = hc/2 - W

Kmax = [ 6.63*10^-34 *3*10⁸ / 330*10^-9 ] - [3.21*10^-19 ]

Kmax = 2.81*10^-19 J

Kmax = 2.81*10^-19 /1.6*10^-19 eV

Kmax = 1.75 eV