A) 510 KeV
B) 931 KeV
C) 510 MeV
D) 931 MeV
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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