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.

Options

A. `0.5 eV`
B. `1.8 eV`
C. `1.02 eV`
D. `2.5 eV`

Answer

C

Explanation

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