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  4. Energy E of a hydrogen atom with principal quantum number n is given by E = (-13.6/n2) 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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Q. 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 :-

AIPMTAIPMT 2004

Solution:

Energy of photon = $E_3 - E_2$
$= - \frac{-13.6}{9} - \left(\frac { - 13.6}{4}\right) = \frac{5}{36} \times 13.6 = 1.9 \,\,eV $