Question

The ground state energy of hydrogen atom is -13.6 ev. What is the ke of an electron in the 3rd excited state

Answer 1

Answer:

3.4 ev

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