On a photosensitive material, when frequency of incident radiation is increased by 30% kinetic energy of emitted photoelectrons increases from `0.4 eV` to `0.9 eV`. The work function of the surface is

Options

A. 1 eV
B. `1.267 eV`
C. `1.4 eV`
D. `1.8 eV`

Answer

B

Explanation

According to the Einstein's equation .
`KE_("max") = hv_(o) - phi_(o)`
Where , `KE_("max") = `maximum kinetic energy and `phi_(0)`
= work function,
Initially , `h = 0.4 + phi_(0) "…"(i)`
When the frequency of incident radiation is increased by 30 % then
`13 hv - 0.9 + phi_(0) "..."(ii)`
Solving Eqs. (i) and (ii), we get
`0.3phi = 0.9 - 1.3 (0.4)`
`:. phi_(o) = (0.38)/(0.3) = 1.267 eV`