Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Hint: Use Einstein's photoelectric equation.

Step 1: Find the work function of the metal.


For the first condition, 

Wavelength of light, λ=600 nm

And for the second condition, 

Wavelength of light, λ'=400 nm

Also, maximum kinetic energy for the second conditions is equal the twice fo the kinetic energy in first condition.

i.e., Kmax=2KmaxHere, Kmax=hcλϕ2Kmax=hcλϕ02(1230600ϕ)=(1230400ϕ) [hc1240 eVnm]ϕ=12301200=1.02 eV