MHT CET · Physics · Dual Nature of Matter
When a certain metal surface is illuminated with light of frequency \(v\), the stopping potential for photoelectric current is \(\mathrm{V}_0\). When the same surface is illuminated by light of frequency \(\frac{v}{2}\), the stopping potential is \(\frac{\mathrm{V}_0}{4}\), the threshold frequency of photoelectric emission is
- A \(\frac{v}{6}\)
- B \(\frac{v}{3}\)
- C \(\frac{2 v}{3}\)
- D \(\frac{4 v}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{v}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{eV}_0=h v-h v_0 ... (i)\\
& \frac{\mathrm{eV}_0}{4}=\frac{h v}{2}-h v_0 ... (ii)
\end{aligned}\)
Dividing equation (i) by equation (ii),
\(\begin{array}{ll}
& 4=\frac{v-v_0}{\frac{v}{2}-v_0} \\
\therefore & 2 v-4 v_0=v-v_0 \\
\therefore & 3 v_0=v \\
\therefore & v_0=\frac{v}{3}
\end{array}\)
& \mathrm{eV}_0=h v-h v_0 ... (i)\\
& \frac{\mathrm{eV}_0}{4}=\frac{h v}{2}-h v_0 ... (ii)
\end{aligned}\)
Dividing equation (i) by equation (ii),
\(\begin{array}{ll}
& 4=\frac{v-v_0}{\frac{v}{2}-v_0} \\
\therefore & 2 v-4 v_0=v-v_0 \\
\therefore & 3 v_0=v \\
\therefore & v_0=\frac{v}{3}
\end{array}\)
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