An electron in the hydrogen atom initially in the fourth excited state makes a transition to nth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be .
Answer
2
Solution
The fourth excited state is ni = 5, where
E5 = - 13.652 = -0.544 eV
so the energy of the final state is En = E5 - 2.86 = -0.544 - 2.86 = -3.404 eV and
- 13.6n2 = -3.404
n2 = 13.63.404 4
n = 2
Answer: 2