Question

Energy E of a hydrogen atom with principal quantum number n is given by E=−13.6n6eV. n The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :-

Answer 1

Answer:

1.9ev

Explanation:

Energy of photon = E3−E2

=  −13.6/9  −  (−13.60)/(4)=5/36×13.6= 1.9eV