1. Use Einstein’s photoelectric equation

For stopping potential $V_s$,

$eV_s = \frac{hc}{\lambda} - \phi$

So for two different wavelengths, the change in stopping potential is

$e\,\Delta V_s = hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$

because the work function $\phi$ cancels out.

2. Given data

• $\lambda_1 = 280\,\text{nm} = 280 \times 10^{-9}\,\text{m}$
• $\lambda_2 = 400\,\text{nm} = 400 \times 10^{-9}\,\text{m}$
• $h = 6.63 \times 10^{-34}\,\text{J s}$
• $c = 3 \times 10^8\,\text{m s}^{-1}$
• $e = 1.6 \times 10^{-19}\,\text{C}$

3. Compute the change in photon energy

$\Delta E = hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$

First,

$hc = (6.63 \times 10^{-34})(3 \times 10^8) = 1.989 \times 10^{-25}\,\text{J m}$

Now,

$\frac{1}{280 \times 10^{-9}} = 3.5714 \times 10^6\,\text{m}^{-1}$

$\frac{1}{400 \times 10^{-9}} = 2.5 \times 10^6\,\text{m}^{-1}$

So,

$\frac{1}{\lambda_1}-\frac{1}{\lambda_2} = (3.5714-2.5)\times 10^6 = 1.0714 \times 10^6\,\text{m}^{-1}$

Hence,

$\Delta E = (1.989 \times 10^{-25})(1.0714 \times 10^6)$

$\Delta E \approx 2.13 \times 10^{-19}\,\text{J}$

4. Convert to change in stopping potential

Since

$e\,\Delta V_s = \Delta E$

$\Delta V_s = \frac{\Delta E}{e} = \frac{2.13 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.33\,\text{V}$

So the stopping potential decreases by about

$\boxed{1.3\,\text{V}}$

when the wavelength is changed from $280\,\text{nm}$ to $400\,\text{nm}$.

5. Check with options

• A: $1.3\,\text{V}$ ✅
• B: $1.1\,\text{V}$
• C: $1.9\,\text{V}$
• D: $0.6\,\text{V}$

Thus, the correct option is A.