Atoms And Nuclei Question 163

Question: The rest energy of an electron is [MP PMT 1996; BCECE 2003]

Options:

A) 510 KeV

B) 931 KeV

C) 510 MeV

D) 931 MeV

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Answer:

Correct Answer: A

Solution:

Rest energy of an electron $ r_{n}\propto n^{2} $

Here $ m_{e}=9.1\times {{10}^{-31}}kg $ and c = velocity of light
$ \therefore $ Rest energy $ =9.1\times {{10}^{-31}}\times {{(3\times 10^{8})}^{2}}joule $

$ =\frac{9.1\times {{10}^{-31}}\times {{(3\times 10^{8})}^{2}}}{1.6\times {{10}^{-19}}}eV=510\ keV $



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