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 emitted photoelectrons depends only on the frequency of the incident light, not on its intensity. This is because a single photon interacts with a single electron, and the energy of the photon is given by \( E = h\nu \). Intensity affects the number of photons (and hence number of electrons ejected), not the energy per photon. \medskip (b) Photoelectric current and intensity: The photoelectric current is proportional to the number of photoelectrons emitted per second. Higher intensity means more photons per unit area per second, resulting in more electrons being ejected (provided the frequency is above the threshold), thus increasing the current. \medskip (c) Stopping potential and frequency: According to Einstein’s photoelectric equation: \[ K_{\text{max}} = h\nu - \phi \] Since \( K_{\text{max}} = eV_0 \), we get: \[ eV_0 = h\nu - \phi \Rightarrow V_0 = \dfrac{h}{e} \nu - \dfrac{\phi}{e} \] This is a linear equation in \( \nu \), which shows that the stopping potential \( V_0 \) varies linearly with frequency for a given surface.