Question

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 n = 3 state to n = 2 state of hydrogen is approximately

• A. 1.9 eV
• B. 1.5 eV
• C. 0.85 eV
• D. 3.4 eV

Answer 1

The correct option is A 1.9 eV

$\Delta E=13.6(\frac{1}{2^2}-\frac{1}{3^2})=13.6\times \frac{5}{36}=1.9eV$