Question

Identify the electronic transition among the consecutive orbits in a hydrogen atom that results in the emission of the electromagnetic relation of the highest frequency. Explain.

Answer 1

E $\propto$ v $\Rightarrow$E=hv

Therefore, the transition from 1st to 2nd orbit will have maximum energy if the transition is in consecutive orbits.

$E_{1\to 2}=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})=13.(1-\frac{1}{4})=10.2eV{E}_{2\to 3}=13.6\left(\frac{5}{36}\right)<10.2eV$