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 wavelength of light used in electron microscope is nearly equal to the resolving power of electron microscope. The de-Broglie wavelength is
$\lambda=\frac{h}{p}=\frac{h}{m v}$
$\Rightarrow v=\frac{h}{m \lambda} \ldots$(i)
Here : $\lambda=10\, pm =10^{-11} m$
$m=9.1 \times 10^{-31} \,kg$
and $h=6.6 \times 10^{-34} \,Js$
So, from e (i), the speed of required electron.
$v=\frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-11}}$
$=7.25 \times 10^{7} m / s$
The energy of electron is $=\frac{1}{2} m v^{2}$
$\left.=\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(7.25 \times 10^{7}\right)^{2} \times \frac{1\, eV }{1.6 \times 10^{-19}}\right]$
$=15 \,keV$