Question

Energy E of 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. 1.5 eV
• B. 0.85 eV
• C. 3.4 eV
• D. 1.9 eV

Answer 1

Answer: (D)

1.9 eV

Answer 2

$E = - \frac{13.6}{n^{2}}eV$

The energy of photon $= E_{3} - E_{2}$

$= \frac{- 13.6}{(3)^{2}} + \frac{13.6}{(2)^{2}} = 13.6\left\lbrack \frac{- 1}{9} + \frac{1}{4} \right\rbrack$

$= 13.6\left\lbrack \frac{- 4 + 9}{36} \right\rbrack = 13.6 \times \frac{5}{36}$

$= 1.9eV.$