Question

A researcher observes hydrogen emitting photons of energy 0.661 eV . What are the quantum numbers of the two states involved in the transition that emits these photons? Enter your answers separated by a comma.

Answer 1

To solve this question draw the energy level diagram of hydrogen atom upto few levels

n=5 E₅ = -0.54 eV

n=4 E₄   = -0.85 eV

n=4 E₃   = -1.5 eV

n=4 E₂ = -3.4 eV

n=4 E₁ = -13.6 eV   

by calculating difference between two energy levels We get energy difference of 0.661 eV arise between n=3 & n=4

E₄ - E₃ =-1.5-0.85 =0.65 ev

so involved energy levels in this transition is n=3,n=4