Question

If an electron in H atom jumps from \(4^{th}\) excited state to \(n^{th}\) energy state, 2.86 eV photon is emitted. The value of ‘n’ will be :-

Answer 1

Answer is : 2

\(2.86 = 13.6 \left(\frac{1}{n^2} - \frac{1}{5^2}\right)\)

\(0.21 = \frac{1}{n^2} - \frac{1}{25}\)

\(n = 2\)