Given below are two statements: one is labell | vedclass

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(B) The stopping potential $V_S$ is given by the Einstein's photoelectric equation: $V_S = \frac{h
u - \phi}{e}$,where $h$ is Planck's constant,$
u$

Vedclass · Accepted Answer

$A$ is false but $R$ is true

Vedclass · Answer

(B) The stopping potential $V_S$ is given by the Einstein's photoelectric equation: $V_S = \frac{h
u - \phi}{e}$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,$\phi$ is the work function,and $e$ is the charge of an electron.From this equation,it is clear that the stopping potential depends only on the frequency of the incident light and the material of the surface (work function). It does not depend on the intensity of the incident light.Therefore,Assertion $A$ is false.The intensity of light is defined as the energy incident per unit area per unit time,which is proportional to the number of photons incident per unit time. Increasing the intensity increases the number of photons,which in turn increases the number of photoelectrons emitted per second,provided the frequency is above the threshold frequency. Thus,Reason $R$ is true.

Similar Questions

Why is photoelectric current proportional to intensity?

Which figure shows the correct variation of applied potential difference $(V)$ with photoelectric current $(I)$ at two different intensities of light $(I_1 < I_2)$ of the same wavelength?

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