1.

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
Answer» E.


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