Question

In a Hydrogen atom, an electron makes a transition from \(n^{th}\) orbit to \(4^{th}\) excited state. Energy released in this transition 0.33 eV find the value of n.

Answer 1

Answer is : 8

\(\Delta E = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\)

\(0.33 = -13.6 \left(\frac{1}{n^2} - \frac{1}{5^2}\right]\)

\(n =8\)