1. Energy lost by the electron

In the Frank–Hertz experiment, the electron initially has energy $E_i = 5.6\ \text{eV}$ and after passing through mercury vapour, it emerges with energy $E_f = 0.7\ \text{eV}$

So the energy lost is $\Delta E = E_i - E_f = 5.6 - 0.7 = 4.9\ \text{eV}$

This energy is used to excite a mercury atom.

2. Photon emitted when atom de-excites

When the excited mercury atom returns to the ground state, it can emit a photon of energy equal to the excitation energy.

Thus, $E_{\gamma} = 4.9\ \text{eV}$

3. Minimum wavelength corresponds to maximum photon energy

The minimum wavelength is given by $\lambda_{\min} = \frac{hc}{E}$

Using $hc \approx 1240\ \text{eV·nm}$

we get $\lambda_{\min} = \frac{1240}{4.9}\ \text{nm}$

$\lambda_{\min} \approx 253\ \text{nm}$

4. Choose the closest option

The closest option is:

• A: $2020\ \text{nm}$
• B: $250\ \text{nm}$
• C: $1700\ \text{nm}$
• D: $220\ \text{nm}$

Hence the correct answer is $\boxed{250\ \text{nm}}$

So, option B is correct.