Question

A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 2.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 6.00 eV. What is the work function of the metal?

Answer 1

`W=2eV`

Step-by-step explanation:

Energy from a light source (photons):

E=h*c/λ

Photoelectric effect and Work function (W):

E=W+Ek

Ek: maximum kinetic energy from photoelectrons

then:

h*c/λ=W+Ek

Case 1:

`h*c/lambda_(1)=W+Ek_(1)` (1)

Case 2:

`h*c/lambda_(2)=W+Ek_(2)`

but
`lambda_(2)=lambda_(1)/2`

`2h*c/lambda_(1)=W+Ek_(2)` (2)

If we divide (2) by (1):

`2=(W+Ek_(2))/(W+Ek_(1))`

`W=Ek_(2)-2Ek{1}=2eV`