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
Show Answer
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 $
