Q162 / 170 · Manipal (MET) 2015 · Physics
Find the frequency of light which ejects electron from a metal surface fully stopped by a retarding potential of 3 V . The photoelectric effect begins in this metal at a frequency of $6 \times 10^{15} \mathrm{~Hz}$.
- Correct answerA$1.324 \times 10^{15} \mathrm{~Hz}$
- B$2.295 \times 10^{16} \mathrm{~Hz}$
- C$3.678 \times 10^{18} \mathrm{~Hz}$
- D$2.7 \times 10^{14} \mathrm{~Hz}$
Solution
According to Einstein's photoelectric equation,
$E=h v-w$
If $V_s$ is retarding or stopping potential and $v_0$ is the threshold frequency, then the above equation becomes
$\begin{aligned} e V_s & =h v-h v_0 \\ h v & =e V_s+h v_0 \\ v & =\frac{e V_s}{h}+v_0\end{aligned}$
Hence, $e=1.6 \times 10^{-19} \mathrm{C}$
$V_s=3 \mathrm{~V}, v_0=6 \times 10^{-14} \mathrm{~Hz}$
Therefore, required frequency,
$\begin{aligned} v & =\frac{1.6 \times 10^{-19} \times 3}{6.63 \times 10^{-34}}+6 \times 10^{14} \\ & =\frac{4.8}{6.63} \times 10^{15}+6 \times 10^{14} \\ & =0.723 \times 10^{15}+6 \times 10^{14} \\ & =7.23 \times 10^{14}+6 \times 10^{14} \\ & =13.23 \times 10^{14} \mathrm{~Hz} \\ & =1.323 \times 10^{15} \mathrm{~Hz}\end{aligned}$
Hence, the correct answer is option: 'A'.