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

For photoelectric emission to take place, the wavelength of incident light should be less than the threshold wavelength for that metal . If `lamda_m` is the threshold wavelength, then work function of metal
`phi = (hc)/lamda_m " or " lamda_m = (hc)/phi`
For metal p, `lamda_(m_p)=(1240 eVnm)/(2.0 eV) = 620 nm`
For metal q, `lamda_(m_p)=(1240 eVnm)/(2.5 eV) = 496 nm`
For metal r, `lamda_(m_p)=(1240 eVnm)/(3.0 eV) = 413.3 nm`
Wavelengths is the incident beam are 550 nm , 450 nm and 350 nm. Thus the photoelectric emission is possible by light of wavelength 350 nm (being shorter than `lamda_m` ) from metals p,q and r. For light of wavelength is possible from metals p and q. For light of wavelength 450 nm `( lt 620 " nm and " 496 "nm")` photoelectric emission is possible from metals p and q. For light of wavelength 550 nm `(lt 620 nm)` , the photoelectric emission is possible from metal p.
where , Intensity = No. of photoelectrons
`:.` I si maximum for p metal , less for q metal and least for r metal . Therefore , the graph (a) is correct.