If the frequency of incident radiation is kep | vedclass

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

Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi

Vedclass · Accepted Answer

remains same.

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident radiation,and $\Phi$ is the work function of the metal.Since the stopping potential $(V_{s})$ is related to the maximum kinetic energy by the relation $eV_{s} = K_{max}$,we have $eV_{s} = h
u - \Phi$.This equation shows that the stopping potential $(V_{s})$ depends only on the frequency $(
u)$ of the incident radiation and the work function $(\Phi)$ of the metal surface.Intensity of light affects the number of photons incident per unit area per unit time,which in turn affects the number of photoelectrons emitted (photoelectric current),but it does not affect the maximum kinetic energy of the individual photoelectrons.Therefore,if the frequency $(
u)$ is kept constant,the stopping potential $(V_{s})$ remains the same regardless of changes in the intensity of the incident light.

If the frequency of incident radiation is kept constant and the experiment is repeated by using incident light of different intensities,then the stopping potential $(V_{s})$

Similar Questions

Assertion: The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.
Reason: The work function of the metal is its characteristic property.

$A$ photocell is illuminated by a small source of light placed at a distance of $1 \ m$. When the same source of light is placed at a distance of $1/2 \ m$,the number of electrons emitted by the photocathode will .......

Threshold frequency for a metal is $15 \times 10^{14} \, Hz$. The light of wavelength $6000 \, \text{Å}$ falls on the metal surface. Then photoelectrons [velocity of light in air, $c = 3 \times 10^8 \, m/s$]:

Light of frequency $1.5 \nu_0$ (where $\nu_0$ is the threshold frequency) is incident on a photosensitive material. If the frequency is halved and the intensity is doubled,what happens to the photocurrent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module