In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $$7.5 \times 10^{-12}$$ m, the minimum electron energy required is close to:

We are told that an electron microscope can resolve details whose size is of the same order as the de-Broglie wavelength of the electrons it uses, that is

$$\lambda \approx 7.5 \times 10^{-12}\ \text{m}.$$

For an electron, the de-Broglie relation gives the momentum in terms of the wavelength:

$$p = \frac{h}{\lambda},$$

where $$h = 6.626 \times 10^{-34}\ \text{J·s}$$ is Planck’s constant.

The kinetic energy of a non-relativistic particle is, by definition,

$$E = \frac{p^{2}}{2m},$$

with $$m = 9.11 \times 10^{-31}\ \text{kg}$$ for the electron mass.

Substituting the momentum expression $$p = \dfrac{h}{\lambda}$$ into the kinetic-energy formula, we obtain

$$E = \frac{1}{2m}\left(\frac{h}{\lambda}\right)^{2}.$$

Now we insert the numerical values step by step:

First compute the square of Planck’s constant:

$$h^{2} = (6.626 \times 10^{-34})^{2} = 4.392 \times 10^{-67}\ \text{J}^{2}\!\cdot\!\text{s}^{2}.$$

Next compute the square of the wavelength:

$$\lambda^{2} = (7.5 \times 10^{-12})^{2} = 5.625 \times 10^{-23}\ \text{m}^{2}.$$

The product in the denominator is

$$2m\lambda^{2} = 2 \times (9.11 \times 10^{-31}) \times (5.625 \times 10^{-23}) = 1.025 \times 10^{-52}\ \text{kg·m}^{2}.$$

We can now evaluate the energy:

$$E = \frac{4.392 \times 10^{-67}}{1.025 \times 10^{-52}} = 4.29 \times 10^{-15}\ \text{J}.$$

To express this energy in electron-volts, we use the conversion

$$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.$$

Hence

$$E = \frac{4.29 \times 10^{-15}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}} \approx 2.68 \times 10^{4}\ \text{eV} = 26.8\ \text{keV}.$$

This value is very close to $$25\ \text{keV}$$ listed among the options.

Hence, the correct answer is Option B.