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Q. The kinetic energy of electron in $ H{{e}^{+}} $ is maximum in:

JIPMERJIPMER 1996

Solution:

We know that, $ {{E}_{n}}=-\frac{13.6\,{{Z}^{2}}}{{{n}^{2}}}eV $ $ {{E}_{n}}\propto \frac{1}{{{n}^{2}}} $ Hence, the energy of electron in $ H{{e}^{+}} $ will be maximum in Ist orbit.