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

To produce photoelectrons, the energy of the light incident should be greater than the work function of the wavelenght of light incident should be shorter than the wavelenght corresponding to the energy or the work function.
Wavelengths less than `lambda_(m)` alon will cause photoeletrons to be ejected.
` lambda_m = (hc)/(phi)`
For p, `lambda_(m_(p)) = (1240 eV nm)/(2.0 eV) = 620` nm
For q, `lambda_(m_(p)) = (1240 eV nm )/(2.5 eV) = 496nm`
For r, `lambda_(m_(r)) = (1240 eV nm)/(3 eV) =413.3` nm
Wavelenghts in the incident beam are 550 nm, 450 nm and 350 nm.
350 nm waves can generate photoelectrons from p, q and r .
450 nm is shorter than `lambda_m` for p and q only and `550 nm lt 620 nm ` only in this group. So it can excite on p - cell Current `prop` intensity.
Intensity = N `h upsilon` of photoelectrons.
`therefore` I is maximum for p cell, one gets the maximum intensity, and next is for q cell and the r cell can give photoelectrons only by 350 nm.
`therefore` I is minimum for r.