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

Energy of wavelength `550nm` is,
`E_(1)=(hc)/(lambda_(1))=(1240 eVnm)/(550nm)=2.25 eV`
Energy wavelength `450nm` is ,
`E_(2)=(1240 eVnm)/(450nm)=2.75 eV`
Energy wavelength `350nm` is ,
`E_(3)=(1240 eVnm)/(350nm)=3.54 eV`
Since light of all the source wavelength are incident on a each plate with equal intensities, the stopping potential will be linked by the light of wavelength having maximum energy i.e., `3.54 eV`. Plate-p is having least work function `phi_(p)(=2.0 eV)`, will have maximum value of stopping potential as it will emit photoelectrons due to light of all the three wavelength of energies `E_(1),E_(2)` and `E_(3)` Therefore, for plate-p saturation current will be maximum and stopping potential will be maximum. Plate- q is having a work function `phi_(0)(=2.5 eV)`, will emit photoelectrons due to light of energy `E_(2)` and `E_(3)`. the value of saturation current and stopping potential will be less than plate-p. Plate-r is having a work function `phi_(r) (=3.0 eV)`, will emit photoelectrons due to light of energy `E_(3)`. its saturation current will be minimum and stopping potential will be least. Thus option (a) is true.