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 $7.5 \times 10^{-12} \mathrm{~m}$, the minimum electron energy required is close to:
Correct Option: 4
(4) Using, $\lambda=\frac{h}{p}\left\{\right.$ given: $\left.\lambda=7.5 \times 10^{-12}\right\}$
$\Rightarrow \mathrm{P}=\frac{\mathrm{h}}{\lambda}$
Minimum energy required,
$\mathrm{KE}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}=\frac{(\mathrm{h} / \lambda)^{2}}{2 \mathrm{~m}}=\frac{\left\{\frac{6.6 \times 10^{-34}}{7.5 \times 10^{-12}}\right\}^{2}}{2 \times 9.1 \times 10^{-31}}$
$\mathrm{J}=25 \mathrm{keV}$
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