Optics Question 110
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 resolved a width of say 10 p.m. If an electron microscope is used, the minimum electron energy required is about
[AIIMS 2004]
Options:
A) 1.5 KeV
B) 15 KeV
C) 150 KeV
D) 1.5 KeV
Show Answer
Answer:
Correct Answer: B
Solution:
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{{{(6.63\times {{10}^{-34}})}^{2}}}{{{(10\times {{10}^{-12}})}^{2}}\times 2\times 9.1\times {{10}^{-31}}}Joule $
$ =\frac{{{(6.63\times {{10}^{-34}})}^{2}}}{{{(10\times {{10}^{-12}})}^{2}}\times 2\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-19}}}eV $ = 15.1 KeV.
BETA
