Photoelectric effect experiments are performed using three different metal plates p, and r having work functions `phi_(P)=2.0eV,phi_(q)=2.5eV` and wavelengths of 550 nm, 450 nm and 350 nm 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

The produce photoelectrons, the enrgy of the light incident hsould be greater than the work function of the wavelength of light incident should be shorter than the wavelength corresponding to the energy or the work function.
Wavelength less than `lambda_("max")` alone will cause photelectrons to be ejected `lambda_(m)=(hc)/(phi)`
For `phi_(p),lambda_(m_(p))=(1240eV nm)/(2.5eV)=496 nm`
For `phi_(r),lambda_(m_(r))=(1240eV nm)/(3eV)=413.3 nm`
Wavelngths 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.
Currept `prop` intensity.
Intensity =Nhv of photoelectrons.
`:.` I is maximum for p cell, one gets the maximum intensity, and next is for q cell and the r cell can give photelectrons only by 350 nm.
`:.` I is minimum for r.