Question

The wavelength of incident light falling on a photosensitive surface is changed from $2000\stackrel{\circ }{A}$ to $2100\stackrel{\circ }{A}$. The corresponding change in stopping potential is

• A. $0.03V$
• B. $0.3V$
• C. $3V$
• D. $3.3V$

Answer 1

Answer: (B)

$0.3V$

Given, ${\lambda }_1=2000\stackrel{\circ }{A}=2000\times {10}^{-10}m$
${\lambda }_1=2\times {10}^{-7}m$
${\lambda }_2=2100\stackrel{\circ }{A}$
${\lambda }_2=2.1\times {10}^{-7}m$
$\therefore \frac{hc}{{\lambda }_1}=W+e{V}_0....\left(i\right)$
and $\frac{hc}{{\lambda }_2}=W+e{V}_0....\left(ii\right)$
Subtracting Eq. (ii) from Eq. (i), we get
$hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=e(V_0-V_0^{\prime })$
Change in stopping potential
$△V=V_0-V_0^{\prime }$
$\Delta V=\frac{hc}{e}(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$
$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}(\frac{1}{2\times {10}^{-7}}-\frac{1}{2.1\times {10}^{-7}})$
$=\frac{6.6\times 3}{1.6}(\frac{1}{2}-\frac{1}{2.1})$
$=\frac{6.6\times 3\times 0.1}{1.6\times 2\times 2.1}=0.3V$.