In hydrogen atom spectrum, when an electron jumps from second excited state to the first excited state, the wavelength of radiation emitted is ' λ '. If the electron jumps from the third excited state to the second orbit, the wavelength of radiation emitted will be (20λ)/(x). The value of x is (a) 18 (b) 27 (c) 21 (d) 36

Answer: (b). Second excited state is n=3, first excited is n=2, third excited is n=4. 3→2: (1)/(λ) ∝ 14-19 = 5/36. 4→2: (1)/(λ') ∝ 14-116 = 3/16. (λ')/(λ) = (5/36)/(3/16) = 80/108 = 20/27 So x = 27, answer (b) 27.

In hydrogen atom spectrum, when an electron jumps from second excited… | MHT CET 2025 (20 Apr, Shift 2)

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Transitions, Photon Emission & Excitation

In hydrogen atom spectrum, when an electron jumps from second excited state to the first excited state, the wavelength of radiation emitted is ' $\lambda$ '. If the electron jumps from the third excited state to the second orbit, the wavelength of radiation emitted will be $\frac{20\lambda}{x}$ . The value of $x$ is

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Answer: (b)

Second excited state is $n=3$ , first excited is $n=2$ , third excited is $n=4$ .
$3\to2$ : $\frac{1}{\lambda} \propto \frac14-\frac19 = \frac{5}{36}$ . $4\to2$ : $\frac{1}{\lambda'} \propto \frac14-\frac1{16} = \frac{3}{16}$ .
$\frac{\lambda'}{\lambda} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}$
So $x = 27$ , answer (b) $27$ .

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