Question

In hydrogen atom, energy of first excited state is –3.4 eV. Find
out KE of the same orbit of Hydrogen atom
(a) + 3.4 eV (b) + 6.8 eV (c) – 13.6 eV (d) + 13.6 eV

Answer 1

The kinetic energy of the same orbit of hydrogen atom is  K.E = + 3.4eV

Option (A) is correct.

Explanation:

Total energy of electron E T  =  Potential energy (P.E) + Kinetic energy (K.E)

For an electron revolving in a circular orbit of radius, r around a nucleus of H atom:

PE = - KZe^2/r

and  

K.E = KZe^2/2r

Thus, E T  = (−KZe^2  / r) + (KZe^2 / 2r) = −KZe^2 / 2r = −K.E

Thus, for ET= -3.4 eV E  

T = −3.4 eV

K.E = + 3.4 eV

Thus the kinetic energy of the same orbit of hydrogen atom is  K.E = +3.4eV

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