Question

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 $6 \times 10^{15} Hz$.

• A. $1.324 \times 10^{15} Hz$
• B. $2.295 \times 10^{16} Hz$
• C. $3.678 \times 10^{18} Hz$
• D. $2.7 \times 10^{14} Hz$

Answer 1

Answer: (A)

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
$\theta V_{s} =h v-h v_{0} $
$h v =\theta V_{s}+h v_{0} $
$v =\frac{\theta V_{g}}{h}+v_{0}$
Hence, $\quad \theta=1.6 \times 10^{-19} C$
$V_{s}=3 V , v_{0}=6 \times 10^{-14} Hz$
Therefore, required frequency,
$ 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} Hz $
$=1.323 \times 10^{16} Hz$