Search any question & find its solution
Question:
Answered & Verified by Expert
When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$, the stopping potential is $\frac{V_0}{4}$. The threshold frequency for photoelectric emission is
Options:
Solution:
2361 Upvotes
Verified Answer
The correct answer is:
$\frac{v}{3}$
We know that,
$e V_0=h v-\phi_0$
where, $\quad V_0=$ stopping potential
and $\quad v=$ frequency of light.
Case II $\frac{e V_0}{4}=\frac{h v}{2}-\phi_0$
From Eqs. (i) and (ii), we get
$\begin{array}{rlrl}
& h v-\phi_0 =2 h v-4 \phi_0 \\
\Rightarrow \quad & h v =3 \phi_0 \Rightarrow h v=3 h v_0 \\
\Rightarrow \quad v_0 & =\frac{h v}{3 h} \Rightarrow v_0=\frac{v}{3}
\end{array}$
$e V_0=h v-\phi_0$
where, $\quad V_0=$ stopping potential
and $\quad v=$ frequency of light.
Case II $\frac{e V_0}{4}=\frac{h v}{2}-\phi_0$
From Eqs. (i) and (ii), we get
$\begin{array}{rlrl}
& h v-\phi_0 =2 h v-4 \phi_0 \\
\Rightarrow \quad & h v =3 \phi_0 \Rightarrow h v=3 h v_0 \\
\Rightarrow \quad v_0 & =\frac{h v}{3 h} \Rightarrow v_0=\frac{v}{3}
\end{array}$
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.