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 :-

AIPMT 2004 Physics and Chemistry Paper

Question : 36 of 100

Marks: +1, -0

Energy E of a hydrogen atom with principal quantum number n is given by E=‌

−13.6

‌eV.
The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately :-

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