A) \[\frac{1}{2}\frac{{{e}^{2}}}{r}\]
B) \[-\frac{{{e}^{2}}}{r}\]
C) \[\frac{m{{e}^{2}}}{r}\]
D) \[-\frac{1}{2}\frac{{{e}^{2}}}{r}\]
Correct Answer: D
Solution :
[d] Total energy of a revolving electron is the sum of its kinetic and potential energy. Total energy \[=K.E.+P.E.\] \[=\frac{{{e}^{2}}}{2r}+\left( -\frac{{{e}^{2}}}{r} \right)=-\frac{{{e}^{2}}}{2r}\]You need to login to perform this action.
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