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Atoms and Nuclei question

2024 · 6 Apr · Shift 1 · Q80

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Atoms and Nuclei
2024 · 6 Apr · Shift 1 · Q80

Atoms and Nuclei question

2024 · 6 Apr · Shift 1 · Q80

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

A
$1: 2$
B
$1: 4$
C
$2: 1$
D
$4: 1$

View written solutionFree

Correct answer: D

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

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

The shortest wavelength in a series occurs when $n_2 \to \infty$ , because then $\frac{1}{\lambda}$ becomes maximum.

Shortest wavelength of Lyman series

For Lyman series, $n_1=1$ .

\frac{1}{\lambda_L}=R\left(1-\frac{1}{\infty^2}\right)=R

So,

\lambda_L=\frac{1}{R}

Shortest wavelength of Balmer series

For Balmer series, $n_1=2$ .

\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=R\cdot \frac{1}{4}

So,

\lambda_B=\frac{4}{R}

Required ratio

\lambda_B : \lambda_L = \frac{4}{R} : \frac{1}{R} = 4:1

Therefore, the ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series is:

\boxed{4:1}

So the correct option is D.

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