Question

A light source of wavelength, (λ), illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.00 eV. A second light source of wavelength (λ/2) ejects photoelectrons with a maximum kinetic energy of 5.60 eV. What is the work function of the metal?

Answer 1

maximum kinetic energy =energy of photon - work function

So, for  light source of wavelength λ, 1.00 eV = ( hc / λ) - work function

work function = ( hc / λ ) - 1.00ev        ------------(1)

For light source of wavelength λ / 2 , 3 e V = ( hc/ ( λ / 2 ) ) - work function

3 eV = 2 hc / λ - work function

= 2 hc / λ - [ ( hc / λ ) - 1.00 ev ]

= ( hc / λ ) + 1.00 eV

So, hc / λ = 5.60 - 1.00 = 4.6 eV

Subctitute this in eq ( 1 ) we get , work fuction = 3.6eV ====================ANSWER)