If the wavelength of the first member of Lyman series of hydrogen is \lambda . The wavelength of the second member will be

Atoms and Nuclei PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q72 with Solution

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Correct answer: C

For hydrogen,

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

For the Lyman series, $n_1 = 1$ .

The first member corresponds to transition:

n_2=2 \to n_1=1

So,

\frac{1}{\lambda_1}=R\left(1-\frac{1}{2^2}\right)=R\left(1-\frac14\right)=\frac{3R}{4}

Given this wavelength is $\lambda$ , so

\lambda_1=\lambda

The second member corresponds to transition:

n_2=3 \to n_1=1

Thus,

\frac{1}{\lambda_2}=R\left(1-\frac{1}{3^2}\right)=R\left(1-\frac19\right)=\frac{8R}{9}

Hence,

\lambda_2=\frac{9}{8R}

Also, from the first member,

\lambda_1=\frac{4}{3R}

So,

\frac{\lambda_2}{\lambda_1}= rac{\frac{9}{8R}}{\frac{4}{3R}}=\frac{9}{8}\cdot\frac{3}{4}=\frac{27}{32}

Therefore,

\lambda_2 = \frac{27}{32}\lambda

The correct option is:

\boxed{\frac{27}{32}\lambda}

which is Option C.