Energy E of a hydrogen atom with principal quantum number n is given by $E=\frac{-13.6}{n^2}$ eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately
Answer & explanation
Correct answer: option 4
$E_{3 \rightarrow 2}=-3.4-(-1.51)=-1.89 eV \Rightarrow \left|E_{3 \rightarrow 2}\right| \approx 1.9 eV$