Search any question & find its solution
Question:
Answered & Verified by Expert
The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_1$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ be the Planck's constant and $e$ be the charge of an electron, then the frequency of light in the second case is
Options:
Solution:
2555 Upvotes
Verified Answer
The correct answer is:
$v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)$
Here, $h v_{1}=\phi_{0}=e V_{1}$ ...(i)
and $h v_{2}=\phi_{0}+e V_{2}$...(ii)
From Eqs. (i) and (ii), we have
$\begin{aligned}
& h\left(v_{2}-v_{1}\right)=e\left(V_{2}-v_{1}\right) \\
\Rightarrow \quad & v_{2}=\frac{e}{h}\left(V_{2}-v_{1}\right)+v_{1}
\end{aligned}$
and $h v_{2}=\phi_{0}+e V_{2}$...(ii)
From Eqs. (i) and (ii), we have
$\begin{aligned}
& h\left(v_{2}-v_{1}\right)=e\left(V_{2}-v_{1}\right) \\
\Rightarrow \quad & v_{2}=\frac{e}{h}\left(V_{2}-v_{1}\right)+v_{1}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.