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Atoms and Nuclei question

2021 · 26 Feb · Shift 1 · Q61

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Atoms and Nuclei
2021 · 26 Feb · Shift 1 · Q61

Atoms and Nuclei question

2021 · 26 Feb · Shift 1 · Q61

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1

If

\lambda

1 and

\lambda

2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of

\lambda

1 :

\lambda

2 is :

A
7 : 135
B
7 : 108
C
1 : 9
D
1 : 3

View written solutionFree

Correct answer: A

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

Find $\lambda_1$ : third member of the Lyman series.

In the Lyman series, transitions end at $n_1 = 1$ .
The members are:
- 1st member: $2 \to 1$
- 2nd member: $3 \to 1$
- 3rd member: $4 \to 1$

So,

\frac{1}{\lambda_1} = R\left(1 - \frac{1}{4^2}\right) = R\left(1 - \frac{1}{16}\right) = R\cdot \frac{15}{16}

Hence,

\lambda_1 = \frac{16}{15R}

Find $\lambda_2$ : first member of the Paschen series.

In the Paschen series, transitions end at $n_1 = 3$ .
The first member is $4 \to 3$ .

So,

\frac{1}{\lambda_2} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right)

Taking LCM $144$ ,

\frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}

Thus,

\frac{1}{\lambda_2} = R\cdot \frac{7}{144} \Rightarrow \lambda_2 = \frac{144}{7R}

Find the ratio $\lambda_1 : \lambda_2$ :

\lambda_1 : \lambda_2 = \frac{16}{15R} : \frac{144}{7R}

Cancel $R$ :

= \frac{16}{15} : \frac{144}{7}

Convert into a simple ratio:

\lambda_1 : \lambda_2 = \frac{16}{15} \cdot \frac{7}{144} = \frac{112}{2160} = \frac{7}{135}

Therefore,

\lambda_1 : \lambda_2 = 7 : 135

Check options:

A: $7:135$ ✅
B: $7:108$ ❌
C: $1:9$ ❌
D: $1:3$ ❌

So the correct option is A.

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