Given that a photon of light of wavelength 10,000 angstrom has an energy equal to `1.23eV`. When light of wavelengths 5000 angstrom and intensity `I_(0)` falls on a photoelectric cell, the saturation current is `0.40 xx 10^(-6)` ampere and the stopping potential is 1.36 volt, then the work function is-

Options

A. `0.43 eV`
B. `1.10eV`
C. `1.36eV`
D. `2.47eV`

Answer

B

Explanation

`E = hv = (hc)/(lambda)`
`:. (E)/(E') = (lambda')/(lambda)` or `(E)/(1.23) = (10,000)/(5,000)`
`:. E = 2.46 eV`
Now `hv - w = (1)/(2)m (w_(max))^(2)`
or `hv - w = eV`,
or `2.46 eV - w = e (1.36)` volt
`:. w = (2.46 - 1.36) eV = 1.1eV`