Question

In an experiment of photoelectric emission, for incident light of $4000\mathring{A},$ the stopping potential is 2V. If the wavelength of incident light is made $3000\mathring{A},$ then stopping potential will be

• A. Less than 2 volt
• B. More than 2 volt
• C. 2 volt
• D. Zero

Answer 1

Answer: (B)

More than 2 volt

The maximum kinetic energy for the photoelectrons is
$E_{max}=h\nu -\varphi$
where, $\nu$ is the frequency of incident light and $\varphi$ is photoelectric work function of metal.

If $V_o$ is the stopping potential then

$e{V}_0=h\frac{c}{\lambda }-\varphi$ .....................(since, $\nu =\frac{c}{\lambda }$)

As per the problem, for incident light of $4000A^{\circ }$, the stopping potential is 2V. When the wavelength of incident light is reduced to $3000A^{\circ }$, then the stopping potential will increase to value more than $2V$(as per the above equation).

So, the answer is option (B).