In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

We have an electron that enters the mercury vapour with a kinetic energy of $$5.6\ \text{eV}$$ and comes out with only $$0.7\ \text{eV}$$.

So the loss in the electron’s energy is

$$\Delta E \;=\; 5.6\ \text{eV} \;-\; 0.7\ \text{eV} \;=\; 4.9\ \text{eV}.$$

In the Frank-Hertz experiment this lost energy is absorbed by a mercury atom, raising the atom from its ground state to its first excited state. When the atom returns to the ground state it emits a photon whose energy is exactly this excitation energy. Hence the photon energy is

$$E_{\text{photon}} \;=\; 4.9\ \text{eV}.$$

For a photon the relation between energy and wavelength is first stated:

$$E \;=\; h \nu \;=\; \frac{h c}{\lambda},$$

where $$h$$ is Planck’s constant and $$c$$ is the speed of light. Solving for the wavelength, we get

$$\lambda \;=\; \frac{h c}{E}.$$

It is convenient to use the product $$h c$$ in the unit “eV·nm”, whose value is well remembered as $$h c \;\approx\; 1240\ \text{eV·nm}$$. Substituting the photon energy,

$$\lambda \;=\; \frac{1240\ \text{eV·nm}}{4.9\ \text{eV}}.$$

Carrying out the division step by step,

$$\lambda \;=\; \frac{1240}{4.9}\ \text{nm} \;=\; 253.06\ \text{nm}\;(\text{approximately}).$$

The closest option to this calculated minimum wavelength is $$250\ \text{nm}$$.

Hence, the correct answer is Option A.