Q64 / 90 · JEE Main 2010 · Physics
Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\max }$. When the ultraviolet light is replaced by $\mathrm{X}-$ rays, both $\mathrm{V}_0$ and $\mathrm{K}_{\max }$ increase.
Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
Of the four choices given after the statements, choose the one that best describes the two statements.
- AStatement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
- BStatement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement- 1.
- CStatement-1 is false, Statement-2 is true.
- Correct answerDStatement-1 is true, Statement- 2 is false.
Solution
Since the frequency of ultraviolet light is less than the frequency of $X$-rays, the energy of each incident photon will be more for X-rays
$K.E_{photoelectron} =h v-\varphi$
Stopping potential is to stop the fastest photoelectron
$$
\mathrm{V}_0=
$$
so, $\mathrm{K}_{\mathrm{E}} \mathrm{E}_{\max }$ and $\mathrm{V}_0$ both increases.
But K.E ranges from zero to K.E $E_{\max }$ because of loss of energy due to subsequent collisions before getting ejected and not due to range of frequencies in the incident light.
Hence, the correct answer is option: 'D'.