Q. 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 resolved a width of say $10\ pm$. If an electron microscope is used, the minimum electron energy required is about
A) 1.5 $keV$
B) 15 $keV$
C) 150 $keV$
D) 1.5 $keV$
Explanation
(b) Wave length of the electron wave be $10 \times 10^{- 12}m,$
Using $\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow E = \frac{h^{2}}{\lambda^{2} \times 2m}$
$= \frac{\left( 6.63 \times 10^{- 34} \right)^{2}}{\left( 10 \times 10^{- 12} \right)^{2} \times 2 \times 9.1 \times 10^{- 31}}Joule$
$= \frac{\left( 6.63 \times 10^{- 34} \right)^{2}}{\left( 10 \times 10^{- 12} \right)^{2} \times 2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19}}eV$
$= 15.1\ KeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.