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. $\left(\frac{h c}{e} = 1240 nm V\right)$

• 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,
$V_{s_{1}}=\frac{1240}{300}-\phi$
where, $\phi$ is the work function of the metal.
The stopping potential for wavelength $400nm$ is given by,
$V_{s_{2}}=\frac{1240}{400}-\phi$
Hence, the difference in stopping potential will be,
$V_{s_{1}}-V_{s_{2}}=\frac{1240}{300}-\frac{1240}{400}$
$\Rightarrow \Delta V_{s}=4.13-3.1$
$\Rightarrow \Delta V_{s}=1.03$
$\Rightarrow \Delta V_{s}\approx1V$