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. none of these

Answer

D

Explanation

`K_(p)=E_(p)-phi_(p)=(1240)/(550)-2.0=0.2545eV`
`K_(q)=E_(q)-phi_(q)=(1240)/(50)-2.5=0.255eV`
`K_(r)=E_(r)-phi_(r)=(1240)/(350)-3.0=0.543eV`
in the above equation, K represents maximum kinetic energy of photoelectrons annd E, the energy of incident right.
From the above values we can see that stopping potential.
`|V_(r)| gt |V_(q)| gt |V_(p)|`
Further, their intensities are equal, but energy of individual photon of r is maximum. hence, number of photons incident (per unit area per unit time) of r can be assumed to be least. hence, saturation currennt of r should be minimu.
keeping these points in mid no option seems to be correct. the correct graph is shown below,
.