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

Explanation

Given,
For the first condition,
Wavelength of light `lamda`= 600 nm and for the second condition,
Wavelength of light `lamda,` = 400 nm
Also maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.
i.e., `" "K_(max)=2K_(max)`
Here, `" "K'_(max)=(hc)/(lamda)-phi`
`implies" "2K_(max)=(hc)/(lamda')-phi_(0)`
`implies" "2((1230)/(600)-phi)=((1230)/(400)-phi)" "[becausehcunderline(~)1240eVnm]`
`implies" "phi=1230/1200=1.02eV`