Question

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

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

Answer 1

Answer: (B)

Given, ${\lambda }_1=2000\mathring{A}=2\times 10^{-7}m$
${\lambda }_2=2100\mathring{A}=2.1\times 10^{-7}m$
$\therefore \frac{hc}{{\lambda }_1}=W+e{V}_0\dots$ (i)
and $\frac{hc}{{\lambda }_2}=W+e{V}_0^{\prime }\dots$ (ii)
Subtracting Eq. (ii) from Eqi (i), we get
$hc\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)=e\left(V_0-V_0{}^{\prime }\right)$
Change in stopping potential
$\Delta V=V_0-V_0^{\prime }=\frac{hc}{e}\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)$
$=\frac{6.6\times 10^{-34}\times 3\times 10^8}{16\times 10^{-19}}\left(\frac{1}{2\times 10^{-7}}-\frac{1}{2.1\times 10^{-7}}\right)$
$=0.3V$