Question

Find the frequency of light which ejects electron from a metal surface fully stopped by a retarding potential of $3V$. 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,
$E=hv-w$
If $V_s$ is retarding or stopping potential and $V_0$ is the threshold frequency, then the above equation becomes
$\theta V_s=hv-h{v}_0$
$hv=\theta V_s+h{v}_0$
$v=\frac{\theta V_g}{h}+v_0$
Hence, $\theta =1.6\times 10^{-19}C$
$V_s=3V,v_0=6\times 10^{-14}Hz$
Therefore, required frequency,
$v=\frac{1.6\times 10^{-19}\times 3}{6.63\times 10^{-34}}+6\times 10^{14}$
$=\frac{4.8}{6.63}\times 10^{15}+6\times 10^{14}$
$=0.723\times 10^{15}+6\times 10^{14}$
$=7.23\times 10^{14}+6\times 10^{14}$
$=13.23\times 10^{14}Hz$
$=1.323\times 10^{16}Hz$