Question

In hydrogen atom energy of first excited state is -3,4 ev then find out the k.E

Answer 1

Answer:

3.4ev

Explanation:

Solution :

Kinetic energy of electron kze22r

Potential energy of electron -kze2r

=> P.E=−2(K.E)

Total energy = P.E +K.E

=−2E.E+K.E

=−KE

Total energy in the first excited state =−3.4ev

KE=−(−3.4ev)

=3.4ev