Question

If in a photoelectric cell, the wavelength of incident light is changed from $4000\mathring{A}$ to $3600\mathring{A},$ then the change in stopping potential will be

• A. 0.34 volt
• B. 0.57 volt
• C. 0.71 volt
• D. 0.23 volt

Answer 1

The correct option is A 0.34 volt

The maximum kinetic energy of photoelectrons in eV is
$E_{max}=h\nu -\varphi$
where,$\nu$ is the frequency of incident light and $\varphi$ is photoelectric work function of metal.
$E_{max}=e{V}_s$
where, $V_s$ is stopping potential and e is electronic charge
$e{V}_s=(h\nu -\varphi )eV$

$e{V}_s=(\frac{hc}{\lambda }-\varphi )eV$

For the wavelength of incident light $4000A^{\circ }$

$e{V}_1=(\frac{hc}{4000}-\varphi )eV$

$e{V}_1=(\frac{12375}{4000}-\varphi )eV$

When the wavelength of incident light changes to $3600A^{\circ }$

$e{V}_2=(\frac{hc}{3600}-\varphi )eV$

$e{V}_2=(\frac{12375}{3600}-\varphi )eV$

therefore

$e{V}_1-e{V}_2=(\frac{12375}{4000}-\frac{\varphi }{e}-\frac{12375}{3600}+\frac{\varphi }{e})eV$

$e(V_1-V_2)=(3.09-3.43)eV=0.34eV$
$V_1-V_2=0.34V$

So, the answer is option (A).