Hello Geetanjali!

I am giving the solution to your question below:-

We have, |K.E|= (1/2 )|P.E|--------(i)

E= (-13.6)× (z^2/n^2) eV/atom

For n= 2, |K.E|= (-13.6)×1/4 = -3.4 eV/atom

Now, |P.E|=2×|K.E|= 2×(-3.4) = -6.8 eV/atom

Hence, the kinetic energy and potential energy of the electron present in the second orbit of Bohr's H-atom is -3.4 eV/atom and -6.8 eV/atom respectively.

Hope you are clear now!

Happy learning:)