Question

When a point light source of power W emitting monochromatic light of wavelength λ is kept at a distance a from a photo-sensitive surface of work function ϕ and area S, we will have:

A. number of photons striking the surface per unit time as $\dfrac{{WS\lambda }}{{4\pi hc{a^2}}}$
B. the maximum energy of the emitted photoelectrons is $\dfrac{1}{\lambda }\left( {hc - \lambda \phi } \right)$
C. the stopping potential needed to stop the most energetic emitted photoelectrons as $e\lambda \left( {hc - \lambda \phi } \right)$
D. photoemission occurs only if λ lies in the range $0 \leqslant \lambda \leqslant \left( {\dfrac{{hc}}{\phi }} \right)$

Answer 1

Hint: In case of photoelectric effect, the cathode is hit by a beam of photons which causes electrons in cathode to come out of the cathode and go to anode. Potential difference is maintained between cathode and anode. The intensity of the incident light determines the current and the energy of the incident light determines the maximum energy of the electron

Formula used:
$W = \dfrac{{nhc}}{{\lambda t}}$

Complete step by step answer:
When radiation is hit on the cathode some of the energy of that radiation will be used by the electron to come out of that electrode which is called a work function and the remaining energy is used as a kinetic energy to reach the anode. When an electron reaches anode automatically current is generated. The amount of current produced depends on the intensity of the radiation.
We can assume that each photon will bring one electron out of the metal. Hence ‘n’ photons will bring ‘n’ electrons out of the metal. The power of the radiation is given by
$W = \dfrac{{nhc}}{{\lambda t}}$
Since it is a point source the light spreads in spherical wavefront form. So for the sphere of surface area $4\pi {a^2}$, the number of photons emitted will be ‘n’. For the metal surface area ‘S’ the number of photons hitting in unit time will be will be
$\eqalign{
  & \Rightarrow W\left( {\dfrac{S}{{4\pi {a^2}}}} \right) = \dfrac{{nhc}}{{\lambda t}} \cr
  & \Rightarrow n = \dfrac{{WS\lambda }}{{4\pi hc{a^2}}} \cr} $
Hence option A will be correct.
Now some energy from the photon is used as a work function to remove electrons from the metal and the rest will be the remaining energy. Hence the maximum energy of an electron will be
$\eqalign{
  & \Rightarrow E = \dfrac{{hc}}{\lambda } - \phi \cr
  & \Rightarrow E = \dfrac{1}{\lambda }\left( {hc - \lambda \phi } \right) \cr} $
Hence option B will be correct
Stopping potential will be
$V = \dfrac{1}{{e\lambda }}\left( {hc - \lambda \phi } \right)$
Hence option C is incorrect
Photoelectric emission occurs only if the energy of the photon is greater than or equal to the work function. That means the wavelength of the photon must be less than or equal to the threshold wavelength
$\eqalign{
  & \Rightarrow \dfrac{{hc}}{\lambda } \geqslant \phi \geqslant 0 \cr
  & \therefore 0 \leqslant \lambda \leqslant \left( {\dfrac{{hc}}{\phi }} \right) \cr} $
Hence option D will be correct

Hence finally A,B,D will be the correct options.

Note:
It doesn’t matter if we increase the number of photons, but as long as the photon energy is less than the required work function, the electron will not emit from the metal. If we apply the stopping potential at the anode then no electron will reach the anode and there will be no photoelectric current.