Photoelectric effect experiments are performed using three different metal plates `p,q` and`r` having work function `phi_(p) = 2.0 eV, phi_(e) = 2.5 eV and phi_(r) = 3.0 eV` respectively A light beam containing wavelength of `550nm , 450 nm` and `350nm ` with equal intensities illuminates each of the plates . The correct `I -V` graph for the experiment is [Take hc = 1240 eV nm]

Options

A.
B.
C.
D.

Answer

A

Explanation

`E_(lambda_(1) = 550 n m) = (1240)/(550)eV = 2.25e V`
`E_(lambda_(2)=450 nm) = (1240)/(450)eV = 2.8 eV`
`E_(lambda_(3) = 350 nm) = (1240)/(350)eV = 3.5 eV`
For metal `r`, only `lambda_(3)` is able to generate photoelectron.
For metal `q`, only `lambda_(2)` and `lambda_(3)` are able to generate photoelectron.
For metal `p`, all wavelength are able to generate photoelectron.
Hence photoelectric current will be maximum for `p` and least for `r`.