This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose…
This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency $v>v_0$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $\mathrm{K}_{\max }$ and $\mathrm{V}_0$ respectively. If the frequency incident on the surface doubled, both the $\mathrm{K}_{\max }$ and $\mathrm{V}_0$ are also doubled.
Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
Statement-1 is false, Statement-2 is true.
Statement-1 is true, Statement-2 is false.
Solution
$
\begin{aligned}
& \mathrm{KE}_{\max }=\mathrm{hu}-\mathrm{hu}_0 \\
& \mathrm{hu}-\mathrm{hu}=\mathrm{e} \times \Delta \mathrm{v} \\
& \mathrm{V}_0=\frac{\mathrm{hu}}{\mathrm{e}}-\frac{\mathrm{hv_{0 }}}{\mathrm{e}}
\end{aligned}
$
' $v$ ' is doubled
$
\begin{aligned}
& \mathrm{KE}_{\max }=2 \mathrm{hu}-\mathrm{hu}_0 \\
& \mathrm{~V}_0^{\prime}=(\Delta \mathrm{V})^{\prime}=\frac{2 \mathrm{hu}}{\mathrm{e}}-\frac{h \mathrm{u}_0}{\mathrm{e}}
\end{aligned}
$
$\frac{\mathrm{KE}_{\max }}{\mathrm{KE}_{\max }}$ may not be equal to 2
$\Rightarrow \frac{\mathrm{V}_0^{\prime}}{\mathrm{V}_0}$ may not equal to 2
$\mathrm{KE} \max =\mathrm{hu}-\mathrm{hv}_0$
$
\mathrm{V}=\frac{h \mathrm{v}}{\mathrm{e}}-\frac{h \mathrm{~h}_0}{\mathrm{e}}
$