question_answer

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         [CBSE PMT 2004]

A)            1.5 eV                                      

B)            0.85 eV

C)            3.4 eV                                      

D)            1.9 eV

Correct Answer: D

Solution :

                   \[{{E}_{3\to 2}}=-3.4-(-1.51)=-1.89\,eV\]Þ \[|{{E}_{3\to 2}}|\ \approx 1.9\,eV\]