1. Use de Broglie wavelength relation

For an electron,

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

To resolve a width of order $7.5\times 10^{-12}\,\text{m}$, we need electron wavelength approximately equal to this value:

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

2. Relate momentum to kinetic energy

Since the expected answer choices are in keV range, first use the non-relativistic approximation:

$K = \frac{p^2}{2m} \quad \Rightarrow \quad p = \sqrt{2mK}$

So,

$\lambda = \frac{h}{\sqrt{2mK}}$

Rearranging,

$K = \frac{h^2}{2m\lambda^2}$

3. Substitute values

Take

$h = 6.626\times 10^{-34}\,\text{J s}, \quad m_e = 9.11\times 10^{-31}\,\text{kg}, \quad \lambda = 7.5\times 10^{-12}\,\text{m}$

Then,

$K = \frac{(6.626\times 10^{-34})^2}{2(9.11\times 10^{-31})(7.5\times 10^{-12})^2}$

First,

$(6.626\times 10^{-34})^2 \approx 4.39\times 10^{-67}$

and

$(7.5\times 10^{-12})^2 = 56.25\times 10^{-24} = 5.625\times 10^{-23}$

So denominator is

$2(9.11\times 10^{-31})(5.625\times 10^{-23})$

$= 1.024875\times 10^{-52}$

Hence,

$K \approx \frac{4.39\times 10^{-67}}{1.024875\times 10^{-52}} \approx 4.28\times 10^{-15}\,\text{J}$

4. Convert into eV

Using

$1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}$

$K \approx \frac{4.28\times 10^{-15}}{1.6\times 10^{-19}} \approx 2.68\times 10^4\,\text{eV}$

$K \approx 26.8\,\text{keV}$

5. Match with options

The closest option is:

$\boxed{25\,\text{keV}}$

So the correct option is A.

6. Check of approximation

Since $26.8\,\text{keV}$ is much smaller than the electron rest energy $511\,\text{keV}$, the non-relativistic approximation is acceptable.

Comparison with stored answer:

Derived answer = A (25 keV)

Stored correct answer = A

They agree.