Question

We wish to see inside an atom. Assuming the atom to have a diameter of $100pm$, this means that one must be able to resolve a width of say $10pm$. 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)

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