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Atoms and Nuclei question

2023 · 8 Apr · Shift 2 · Q62

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Atoms and Nuclei
2023 · 8 Apr · Shift 2 · Q62

Atoms and Nuclei question

2023 · 8 Apr · Shift 2 · Q62

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1

The ratio of wavelength of spectral lines

\mathrm{H}_{\alpha}

and

\mathrm{H}_{\beta}

in the Balmer series is

\frac{x}{20}

. The value of

x

is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 27

Use the Balmer series formula

For hydrogen spectral lines in the Balmer series,

\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), \quad n=3,4,5,\dots

Here:

$H_{\alpha}$ corresponds to transition $n=3 \to 2$
$H_{\beta}$ corresponds to transition $n=4 \to 2$

Find wavelength of $H_{\alpha}$

For $n=3$ ,

\frac{1}{\lambda_{\alpha}}=R\left(\frac{1}{4}-\frac{1}{9}\right) =R\left(\frac{9-4}{36}\right) =\frac{5R}{36}

So,

\lambda_{\alpha}=\frac{36}{5R}

Find wavelength of $H_{\beta}$

For $n=4$ ,

\frac{1}{\lambda_{\beta}}=R\left(\frac{1}{4}-\frac{1}{16}\right) =R\left(\frac{4-1}{16}\right) =\frac{3R}{16}

So,

\lambda_{\beta}=\frac{16}{3R}

Take the ratio

\frac{\lambda_{\alpha}}{\lambda_{\beta}}=\frac{\frac{36}{5R}}{\frac{16}{3R}} =\frac{36}{5R}\cdot\frac{3R}{16} =\frac{108}{80} =\frac{27}{20}

Given,

\frac{\lambda_{H_{\alpha}}}{\lambda_{H_{\beta}}}=\frac{x}{20}

Comparing,

\frac{x}{20}=\frac{27}{20} \Rightarrow x=27

Final Answer

\boxed{27}

The derived answer matches the stored correct answer.

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