On a photosensitive surface, if the intensity | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The stopping potential $(V_0)$ is determined by the maximum kinetic energy of the emitted photoelectrons, which is given by Einstein's photoelectr

Vedclass · Accepted Answer

remains unchanged

Vedclass · Answer

(D) The stopping potential $(V_0)$ is determined by the maximum kinetic energy of the emitted photoelectrons, which is given by Einstein's photoelectric equation: $K_{max} = h
u - \Phi = eV_0$. Here, $h$ is Planck's constant, $
u$ is the frequency of incident radiation, $\Phi$ is the work function of the surface, and $e$ is the charge of an electron. Since the stopping potential depends only on the frequency of the incident radiation and the nature of the material (work function), it is independent of the intensity of the incident radiation. Therefore, increasing the intensity of the incident radiation does not change the stopping potential.

On a photosensitive surface, if the intensity of incident radiation is increased, the stopping potential

Similar Questions

For most metals, threshold frequencies lie in which region?

For a given source,with an increase in intensity of incident radiation,the . . . . . .

The electromagnetic theory of light failed to explain:

When a monochromatic point source of light is at a distance of $0.2\, m$ from a photoelectric cell,the cut-off voltage and the saturation current are respectively $0.6\, V$ and $18.0\, mA$. If the same source is placed $0.6\, m$ away from the photoelectric cell,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module