1. Use Einstein’s photoelectric equation

For a given photo-sensitive material, $K_{\max}=h\nu-\phi$ and the stopping potential is $eV_s=K_{\max}$ So, $V_s=\frac{hc}{e\lambda}-\frac{\phi}{e}$

For the same material, the work function term remains constant. Hence for two wavelengths $\lambda_1$ and $\lambda_2$, $V_{s2}-V_{s1}=\frac{hc}{e}\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)$

Using $\frac{hc}{e}=1240\ \text{V·nm}$

2. Given data

• First wavelength: $\lambda_1=670.5\ \text{nm}$
• First stopping voltage: $V_{s1}=0.48\ \text{V}$
• Second wavelength: $\lambda_2=474.6\ \text{nm}$

We need to find $V_{s2}$.

3. Compute the change in stopping potential

$V_{s2}=V_{s1}+1240\left(\frac{1}{474.6}-\frac{1}{670.5}\right)$

Now, $\frac{1}{474.6}\approx 0.002107$ $\frac{1}{670.5}\approx 0.001491$

So, $\frac{1}{474.6}-\frac{1}{670.5}\approx 0.000616$

Multiply by $1240$: $1240\times 0.000616\approx 0.764$

Thus, $V_{s2}=0.48+0.764=1.244\ \text{V}$

$V_{s2}\approx 1.25\ \text{V}$

4. Check options

• A: $0.96\ \text{V}$
• B: $1.25\ \text{V}$
• C: $0.24\ \text{V}$
• D: $1.5\ \text{V}$

Hence, the correct option is B.

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.