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.



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