Question

Answer the following giving reasons:
(a) The maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
(b) Photoelectric current increases with the increase in the intensity of the incident radiation.
(c) The stopping potential \( V_0 \) varies linearly with the frequency \( \nu \) of the incident radiation for a given photosensitive surface.

Hint:

Remember: - Intensity affects the number of electrons (current), - Frequency affects the energy of electrons (kinetic energy), - Stopping potential is a direct measure of electron energy and hence linked to frequency, not intensity.

Answer 1

(a) Maximum kinetic energy and intensity: The maximum kinetic energy of ejected photoelectrons is determined solely by the incident light's frequency, not its intensity. This is due to the one-to-one interaction between a photon and an electron, where the photon's energy is \( E = hu \). Intensity influences the quantity of photons (and consequently, the number of ejected electrons), but not the energy carried by each photon.\medskip(b) Photoelectric current and intensity: The photoelectric current directly correlates with the rate of photoelectron emission. Increased intensity signifies a greater flux of photons per unit area per second, leading to more electrons being ejected (provided the frequency exceeds the threshold), thereby augmenting the current.\medskip(c) Stopping potential and frequency: Based on Einstein’s photoelectric equation:\[K_{\text{max}} = hu - \phi\]Substituting \( K_{\text{max}} = eV_0 \), we obtain:\[eV_0 = hu - \phi \Rightarrow V_0 = \dfrac{h}{e} u - \dfrac{\phi}{e}\]This equation, being linear in \( u \), demonstrates that for a specific surface, the stopping potential \( V_0 \) exhibits a linear relationship with frequency.