Question

A stopping potential of 0.82 volt is required to stop the photo electrons emitted from a metallic surface by light of wavelength $4000\mathring{A}.$ The stopping potential for wavelength $3000\mathring{A}$ will be

• A. 1.1 volt
• B. 1.85 volt
• C. 2.03 volt
• D. 2.5 volt

Answer 1

The correct option is B 1.85 volt

Here, a stopping potential of 0.82 volt is required to stop the photo electrons emitted from a metallic surface by light of wavelength $4000A^{\circ }$, hence
$eV=h\nu -\varphi$
where, V is the stopping potential, e is the electronic charge, $\nu$ is the frequency of incident light and $\varphi$ is photoelectric work function of metal, h is Plank's constant.

$eV=h\frac{c}{\lambda }-\varphi$

$1.6\times 0.82\times {10}^{-19}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}-\varphi$

$1.312\times {10}^{-19}=4.95\times {10}^{-19}-\varphi$
$\varphi =(4.95-1.312)\times {10}^{-19}$
$\varphi =3.638eV$
The stopping potential for wavelength $3000A^{\circ }$

$V=\frac{h\frac{c}{\lambda }-\varphi }{e}$

$V=\frac{\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-10}}-3.638\times {10}^{-19}}{1.6\times {10}^{-19}}$

$V=\frac{2.962}{1.6}$

$V=1.85V$

So, the answer is option (B).