A) four times of its P.E.
B) twice of its P. E.
C) equal to its P. E.
D) half of its P. E.
Correct Answer: D
Solution :
K.E. of electron\[=\frac{13.6{{Z}^{2}}}{{{n}^{2}}}eV\] P.E. of electron\[=\frac{2\times 13.6{{Z}^{2}}}{{{n}^{2}}}eV\] \[\therefore \]K.E. of electron\[=\frac{1}{2}\times \]P.E. of electron
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