The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are :

Structure of Atom PYQ — JEE Main 2019 · 10 Apr · Shift 2 · Q12 with Solution

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Correct answer: C

Use the Rydberg formula for hydrogen spectral lines:

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

For a given spectral series, $n_1$ is fixed:

Lyman: $n_1 = 1$
Balmer: $n_1 = 2$
Paschen: $n_1 = 3$
Brackett: $n_1 = 4$
Pfund: $n_1 = 5$

Find the shortest wavelength in a series.

The shortest wavelength means maximum energy transition, which occurs when:

n_2 \to \infty

So,

\frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_1^2} - 0\right) = \frac{R}{n_1^2}

Hence,

\lambda_{\min} = \frac{n_1^2}{R}

Thus shortest wavelength is proportional to $n_1^2$ :

\lambda_{\min} \propto n_1^2

Required ratio is about 9.

So for two series with lower levels $n_a$ and $n_b$ ,

\frac{\lambda_{\min,1}}{\lambda_{\min,2}} = \frac{n_a^2}{n_b^2}

We need this ratio to be approximately $9$ .

That means:

\frac{n_a^2}{n_b^2} = 9 \implies \frac{n_a}{n_b} = 3

Among the series values $1,2,3,4,5$ , the pair with ratio $3$ is $3:1$ .

That corresponds to:

Paschen: $n_1 = 3$
Lyman: $n_1 = 1$

So,

\frac{\lambda_{\min}(\text{Paschen})}{\lambda_{\min}(\text{Lyman})} = \frac{3^2}{1^2} = 9

Check options:

A: Paschen and Pfund $\frac{\lambda_{\min}(P)}{\lambda_{\min}(Pf)} = \frac{9}{25} \neq 9$
B: Balmer and Brackett $\frac{4}{16} = \frac{1}{4} \neq 9$
C: Lyman and Paschen Ratio of shortest wavelengths is either $\frac{1}{9}$ or $9$ depending on order. Since question says ratio is about 9, the pair is correct.
D: Brackett and Pfund $\frac{16}{25} \neq 9$

Therefore, the correct pair is Lyman and Paschen.

Final Answer:

\boxed{\text{C: Lyman and Paschen}}

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View written solutionFree

Correct answer: C

Use the Rydberg formula for hydrogen spectral lines:

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

For a given spectral series, $n_1$ is fixed:

Lyman: $n_1 = 1$
Balmer: $n_1 = 2$
Paschen: $n_1 = 3$
Brackett: $n_1 = 4$
Pfund: $n_1 = 5$

Find the shortest wavelength in a series.

The shortest wavelength means maximum energy transition, which occurs when:

n_2 \to \infty

So,

\frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_1^2} - 0\right) = \frac{R}{n_1^2}

Hence,

\lambda_{\min} = \frac{n_1^2}{R}

Thus shortest wavelength is proportional to $n_1^2$ :

\lambda_{\min} \propto n_1^2

Required ratio is about 9.

So for two series with lower levels $n_a$ and $n_b$ ,

\frac{\lambda_{\min,1}}{\lambda_{\min,2}} = \frac{n_a^2}{n_b^2}

We need this ratio to be approximately $9$ .

That means:

\frac{n_a^2}{n_b^2} = 9 \implies \frac{n_a}{n_b} = 3

Among the series values $1,2,3,4,5$ , the pair with ratio $3$ is $3:1$ .

That corresponds to:

Paschen: $n_1 = 3$
Lyman: $n_1 = 1$

So,

\frac{\lambda_{\min}(\text{Paschen})}{\lambda_{\min}(\text{Lyman})} = \frac{3^2}{1^2} = 9

Check options:

A: Paschen and Pfund $\frac{\lambda_{\min}(P)}{\lambda_{\min}(Pf)} = \frac{9}{25} \neq 9$
B: Balmer and Brackett $\frac{4}{16} = \frac{1}{4} \neq 9$
C: Lyman and Paschen Ratio of shortest wavelengths is either $\frac{1}{9}$ or $9$ depending on order. Since question says ratio is about 9, the pair is correct.
D: Brackett and Pfund $\frac{16}{25} \neq 9$

Therefore, the correct pair is Lyman and Paschen.

Final Answer:

\boxed{\text{C: Lyman and Paschen}}

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