Question

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

• A. 15 keV
• B. 1.5 keV
• C. 150 keV
• D. 1.5 MeV

Answer 1

Answer: (A)

15 keV

Answer 2

The de-Broglie wavelength $(\lambda)$ is given by
$\lambda=\frac{h}{p}=\frac{h}{m v}$
where $h$ is Planck's constant, $m$ the mass and $v$ the velocity.
Given, $\lambda=10\, pm =10^{-11} m$,
$m =9.1 \times 10^{-31} kg ,$
$h =6.6 \times 10^{-34} J - s .$
$\therefore v =\frac{h}{m \lambda}=\frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-11}}$
$\Rightarrow v =7.25 \times 10^{7} m / s$
Also, kinetic energy is the energy possessed due to velocity $(v)$ is given by
$KE =\frac{1}{2} m v^{2}$
Given, $m=9.1 \times 10^{-31} kg$,
$v =7.25 \times 10^{7} m / s$
$\therefore KE =\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(7.25 \times 10^{7}\right)^{2}$
Since, $1\, eV =1.6 \times 10^{-19} J$
$\therefore KE =\frac{1}{2} \times 9.1 \times 10^{-31} \times \frac{\left(7.25 \times 10^{7}\right)^{2}}{1.6 \times 10^{-19}}$
$=15 \,keV$