Question

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300nm$ to $400nm$ . Choose the closest value of change in the stopping potential from the given options. $\left(\frac{hc}{e}=1240nmV\right)$

• A. $2.0V$
• B. $0.5V$
• C. $1.0V$
• D. $1.5V$

Answer 1

Answer: (C)

The stopping potential for the wavelength $300nm$ is given by,
$V_{s_1}=\frac{1240}{300}-\varphi$
where, $\varphi$ is the work function of the metal.
The stopping potential for wavelength $400nm$ is given by,
$V_{s_2}=\frac{1240}{400}-\varphi$
Hence, the difference in stopping potential will be,
$V_{s_1}-V_{s_2}=\frac{1240}{300}-\frac{1240}{400}$
$\Rightarrow \Delta V_s=4.13-3.1$
$\Rightarrow \Delta V_s=1.03$
$\Rightarrow \Delta V_s\approx 1V$