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# We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the minimum electron energy required is about

$\begin {array} {1 1} (a)\;1.5\: KeV & \quad (b)\;15\: KeV \\ (c)\;150\: KeV & \quad (d)\;1.5\: MeV \end {array}$

$E = \large\frac{1}{2} mv^2 = \large\frac{1}{2} m \bigg(\large\frac{h}{m\lambda} \bigg)^2\: \: \: \: \bigg[ \lambda= \large\frac{h}{mv} \bigg]$
$= \large\frac{1}{2} \large\frac{h^2}{2m \lambda^2}$
$E \: in \: eV = \large\frac{1}{2} \large\frac{h^2}{2mλ^2e}$
$= \large\frac{(6.63 \times 10^{-34} )^2}{ (2 \times 9.11 \times 10^{-31} \times 10^{-11} \times 10^{-11} \times 1.6 \times 10^{-19} )}$
$= 1.5 \times 10^4 \: eV= 15\: keV$