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Atoms and Nuclei question

2025 · 7 Apr · Shift 1 · Q67

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Atoms and Nuclei
2025 · 7 Apr · Shift 1 · Q67

Atoms and Nuclei question

2025 · 7 Apr · Shift 1 · Q67

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

A
$5: 27$
B
$27: 5$
C
$3: 4$
D
$5: 36$

View written solutionFree

Correct answer: A

Use the hydrogen spectral formula

For hydrogen, the wavelength of a spectral line is given by

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

where $R$ is the Rydberg constant.

Largest wavelength in a series

In a given series, $n_1$ is fixed. The largest wavelength corresponds to the smallest energy difference, i.e. the smallest possible value of $n_2$ .

So:

For Lyman series, $n_1=1$ , hence largest wavelength occurs for $n_2=2$ .
For Balmer series, $n_1=2$ , hence largest wavelength occurs for $n_2=3$ .

Largest wavelength of Lyman series

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

Thus,

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

Largest wavelength of Balmer series

\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac14-\frac19\right)=R\left(\frac{5}{36}\right)

Thus,

\lambda_B=\frac{36}{5R}

Find the ratio

\lambda_L : \lambda_B = \frac{4}{3R} : \frac{36}{5R}

Cancel $R$ :

= \frac{4}{3} : \frac{36}{5}

Multiply both by $15$ :

= 20 : 108 = 5 : 27

Conclusion

The required ratio is

\boxed{5:27}

So the correct option is A.

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