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In hydrogen atom, energy offirst excited state is -3.4 eV. Then find out K.E. of same orbit of hydrogen atom 

(a) +3.4 eV 

(b) +6.8 eV 

(c) -13.6 eV 

(d) +13.6 eV

1 Answer

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Best answer

The correct option is : (a) +3.4 eV 

Explanation :

. .. Kinetic energy = -En 

Energy of first excited state is -3.4 eV

. .. Kinetic energy of same orbit (n = 2) will be + 3.4 eV.

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