Question

If we plot a graph of stopping potential of photoelectrons against the frequency (in hertz) of radiation that has emitted, then its slope would be:
($h=6.6\times {10}^{-34})$

• A. 6.625 $\times {10}^{-34}$
• B. 0.414 $\times {10}^{-14}$
• C. 3.656 $\times {10}^{12}$
• D. 5.796 $\times {10}^{18}$

Answer 1

The correct option is B 0.414 $\times {10}^{-14}$

According to Einstein Photoelectric equation,
$h\nu$ = $h{\nu }_{\circ }$ + $e{V}_{\circ }$
Where,
$V_o=\text{Stopping potential}$
${\nu }_o=\text{Threshold frequency}$
$\nu =\text{Frequency of incident radiation}$
Writing this equation in terms of $y=mx+c$, we get
$V_{\circ }$ = $\frac{h(\nu -{\nu }_{\circ })}{e}$
On comparing this equation with $y=mx+c$, we get
$\text{Slope}=m=\frac{h}{e}$
Therefore, $m=\frac{h}{e}$ = $\frac{6.6\times {10}^{-34}}{1.6\times {10}^{-19}}$
$=4.125\times$ ${10}^{-15}$
$=0.413\times$ ${10}^{-14}$
Hence, option (b) is correct.