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

• A. 0. 85 eV
• B. 3.4 eV
• C. 1.9 eV
• D. 1.5 eV

Answer 1

The correct option is C 1.9 eV

$\Delta E=E_3-E_2$

$=\frac{-13.6}{3^2}+\frac{13.6}{2^2}$

$=13.6[\frac{1}{4}-\frac{1}{9}]$

$eV=1.9eV$