A hydrogen-like species can emit a maximum energy photon of 204 eV. It is in an excited state of quantum number 2n. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. What is the value of n ?

Answer

2

Explanation

`204 = 13.6 xx 7^(2) (1- 1/(4n^(2)))`
`40.8 = 13.6 xx 7^(2) xx (1/n^(2) - 1/(4n^(2)))`
On dividing we get,
`15 = 4n^(2) -1`
n = 2 (approx).