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NEET 2004 (AIPMT) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from NEET 2004 (AIPMT)

Q77NEET 2004 · AIPMTEasy

Transitions, Photon Emission & Excitation

Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^{2}}\text{ eV}$ . 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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Answer: (d)

$E = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ eV
$E = 13.6 \times \frac{5}{36} \approx 1.89$ eV
So the answer is (d) $1.9$ eV.

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