A certain metal has a work function of `Phi = 2` e V . It is irradiated first with 1 W of 400 nm light and later with 1 W of 800 nm light . Among the following , the correct statement is :
[Given : Planck constant (h) = `6.626 xx 10^(-34) m^(2) kg s^(-1)` , Speed of light (e) = `3 xx 10^(8) m s^(-1)]`

Options

A. Both colors of light give rise to same number of photoelectrons.
B. 400 nm light gives rise to less energetic photoelectrons than 800 nm light.
C. 400 nm light leads to more photoelectrons.
D. 800 nm light leads to more photoelectrons.

Answer

C

Explanation

Work function of metal (`phi`) = 2 eV
Energy of photon (`lamda` = 400 nm) = `(hc)/lamda=3.105eV`
Energy of photon (`lamda` = 800 nm) = `(hc)/(lamda)=1.5525eV`
Hence, photon with `lamda` = 400 nm will emit photoelectrons while photon with `lamda` = 800 nm will not emit photoelectrons.