If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be (a) 16/25λ (b) 9/16λ (c) 20/7λ (d) 20/13λ

Answer: (c). (1)/(λ) ∝ ((1)/(nf²)-(1)/(ni²)) 3→2: 5/36. 4→3: 19-116 = 7/144. λ' = λ × (5/36)/(7/144) = 20/7λ So the answer is (c) 20/7λ.

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd… | NEET 2016 (Phase 2)

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Transitions, Photon Emission & Excitation

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$ . When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

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Answer: (c)

$\frac{1}{\lambda} \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
$3\to2$ : $\frac{5}{36}$ . $4\to3$ : $\frac19-\frac1{16} = \frac{7}{144}$ .
$\lambda' = \lambda \times \frac{5/36}{7/144} = \frac{20}{7}\lambda$
So the answer is (c) $\frac{20}{7}\lambda$ .

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