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, $\lambda_{1}=2000\,\mathring{A}=2 \times 10^{-7} m$
$\lambda_{2}=2100\,\mathring{A}=2.1 \times 10^{-7} m$
$\therefore \frac{h c}{\lambda_{1}}=W+e V_{0} \ldots$ (i)
and $\frac{h c}{\lambda_{2}}=W+e V_{0}' \ldots$ (ii)
Subtracting Eq. (ii) from Eqi (i), we get
$h c\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=e\left(V_{0}-V_{0}{ }'\right)$
Change in stopping potential
$\Delta V=V_{0}-V_{0}'=\frac{h c}{e}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$
$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{16 \times 10^{-19}}\left(\frac{1}{2 \times 10^{-7}}-\frac{1}{2.1 \times 10^{-7}}\right)$
$=0.3\, V$