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Q. The kinetic energy of an electron revolving around a nucleus will be

Punjab PMETPunjab PMET 2005Atoms

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