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

Answer: (D)

3.4 Ev

Answer 2

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