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

Question

Dr. Abubacker · Accepted Answer

(d) $E_{3 \rightarrow 2} = - 1.51 - ( - 3.4) = 1.89eV$
$\Rightarrow \left| E_{3 \rightarrow 2} \right| = 1.9eV$

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

Explanation

Reviewed by Dr. Abubacker

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