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NEET 1992 (AIPMT) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from NEET 1992 (AIPMT)

Q82NEET 1992 · AIPMTEasy

Transitions, Photon Emission & Excitation

The ionization energy of hydrogen atom is 13.6 eV. Following Bohr's theory, the energy corresponding to a transition between $3$ rd and $4$ th orbit is

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Answer: (d)

$E_n = -\frac{13.6}{n^2}$ eV, so $\Delta E = 13.6\left(\frac19-\frac1{16}\right)$
$\Delta E = 13.6 \times \frac{7}{144} \approx 0.66$ eV
So the answer is (d) $0.66$ eV.

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