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MCQEasyJEE 2024Bohr's Model & Hydrogen Spectrum

JEE Physics 2024 Question with Solution

If the wavelength of the first member of the Lyman series of hydrogen is λ\lambda, the wavelength of the second member will be:

  • A

    (2732)λ\left(\frac{27}{32}\right)\lambda

  • B

    (3227)λ\left(\frac{32}{27}\right)\lambda

  • C

    (275)λ\left(\frac{27}{5}\right)\lambda

  • D

    (527)λ\left(\frac{5}{27}\right)\lambda

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The wavelength of the first member of the Lyman series is λ\lambda.

Find: The wavelength of the second member of the Lyman series.

For the Lyman series, the electron falls to n1=1n_1 = 1 from higher energy levels. Use the Rydberg formula:

1λ=R(1n12−1n22)\frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)

For the first member of the Lyman series, the transition is n2=2n_2 = 2 to n1=1n_1 = 1:

1λ=R(1−14)=3R4\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}

For the second member, the transition is n2=3n_2 = 3 to n1=1n_1 = 1. Let its wavelength be λ2\lambda_2:

1λ2=R(1−19)=8R9\frac{1}{\lambda_2} = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9}

Now compare the two expressions:

1λ1λ2=3R48R9\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_2}} = \frac{\frac{3R}{4}}{\frac{8R}{9}}

So,

λ2λ=3489−1=2732\frac{\lambda_2}{\lambda} = \frac{\frac{3}{4}}{\frac{8}{9}}^{-1} = \frac{27}{32}

Hence,

λ2=2732λ\lambda_2 = \frac{27}{32}\lambda

Therefore, the wavelength of the second member is 2732λ\frac{27}{32}\lambda. The correct option is A.

Ratio Method

Given: First-member wavelength is λ\lambda.

Find: Second-member wavelength in terms of λ\lambda.

For the Lyman series, 1λ∝(1−1n2)\frac{1}{\lambda} \propto \left(1-\frac{1}{n^2}\right) because the lower level is fixed at n=1n=1.

So for the first and second members:

1λ∝1−122=34\frac{1}{\lambda} \propto 1-\frac{1}{2^2} = \frac{3}{4} 1λ′∝1−132=89\frac{1}{\lambda'} \propto 1-\frac{1}{3^2} = \frac{8}{9}

Hence,

λ′λ=3489=2732\frac{\lambda'}{\lambda} = \frac{\frac{3}{4}}{\frac{8}{9}} = \frac{27}{32}

Therefore,

λ′=2732λ\lambda' = \frac{27}{32}\lambda

Thus, the correct option is A.

Common mistakes

  • Using the second member as transition n=2→1n=2 \to 1 is incorrect. That is the first member of the Lyman series. The second member is n=3→1n=3 \to 1.

  • Reversing the wavelength ratio is a common error. Since 1λ\frac{1}{\lambda} is proportional to the transition factor, the wavelength itself varies inversely with that factor.

  • Confusing the Lyman series with Balmer series is incorrect. In the Lyman series, the final level is always n1=1n_1=1, not n1=2n_1=2.

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