Question

In a photoelectric effect measurement, the stopping potential for a given metal is found to be $V_0$ volt when radiation of wavelength ${\lambda }_0$ is used. If radiation of wavelength $2{\lambda }_0$ is used with the same metal than the stopping potential (in volt) will be:

• A. $\frac{V_0}{2}$
• B. $2V_0$
• C. $V_0+\frac{hc}{2e{\lambda }_0}$
• D. $V_0-\frac{hc}{2e{\lambda }_0}$

Answer 1

The correct option is D $V_0-\frac{hc}{2e{\lambda }_0}$

Photoelectric effect
$K{E}_{max}=\frac{hc}{\lambda }-W_o$
Stopping potential
$V_{s}e=\frac{hc}{\lambda }-W_o$
given
$V_{o}e=\frac{hc}{{\lambda }_o}-W_o$ -------------(1)
then for double the wavelength
$V_{s}e=\frac{hc}{2{\lambda }_o}-W_o$ ----------------(2)
Substracting (1) - (2)
$(V_o-V_s)e=\frac{hc}{2{\lambda }_o}$
$V_s=V_o-\frac{hc}{2e{\lambda }_o}$