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)

$1.324\times {10}^{15}Hz$

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

$e{V}_s=hv-h{v}_0$

$hv=e{V}_s+h{v}_0$

$v=\frac{e{V}_s}{h}+v_0$

Hence, $e=1.6\times {10}^{-19}C$

$V_s=3V$, $v_0=6\times {10}^{-14}Hz$

Therefore, required frequency,

$v=\frac{1.6\times {10}^{-19}\times 3}{6.63\times {10}^{-34}}+6\times {10}^{14}$

$=\frac{4.8}{6.63}\times {10}^{16}+6\times {10}^{14}$

$=0.723\times {10}^{15}+6\times {10}^{14}$

$=7.23\times {10}^{14}+6\times {10}^{14}Hz$

$=13.23\times {10}^{14}Hz$

$=1.323\times {10}^{15}Hz$