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Atomic Structure

Section: Chemistry

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Question : 10 of 32

Marks: +1, -0

The wavenumber of the first line

(n_2=3)

in the Balmer series of hydrogen is

\overline{v}_1 \text{ cm}^{-1}

. What is the wavenumber (in

\text{cm}^{-1}

) of the second line (

n_2=4

) in the Balmer series of

\mathrm{He}^{+}

[AP EAPCET 22 May 2025 S1]

$\frac{5 \overline{v}_1}{27}$
$\frac{27 \overline{v}_1}{5}$
$\frac{27 \overline{v}_1}{20}$
$\frac{20 \overline{v}_1}{27}$

Solution:

Using the Rydberg formula for wavenumber

\overline{v}=R z^2 \left[\;\frac{1}{n_1^2}-\;\frac{1}{n_2^2}\right]

for wavenumber of 1st line of H atom

n_1=2,\ n_2\;=3,\ z=1

\overline{v}_{H}\;=R \left[\;\frac{1}{4}-\;\frac{1}{9}\right]

\overline{v}_{H}\;=R \left[\;\frac{5}{36}\right] \dots (i)

The wavenumber of 2 nd line of

\mathrm{He}^{+}

\;n_1=2,\ n_2=4,\ z=2

\; \overline{v}_{\mathrm{He}}^{+}=4 R \left[\;\frac{1}{4}-\;\frac{1}{16}\right] =R \left(\;\frac{3}{4}\right) \dots (ii)

Divide Eq. (ii) by Eq. (i)

\overline{v}_{\mathrm{He}}^{+} = \;\frac{27}{5} \overline{v}_{H} \;\text{ or }\; \;\frac{27}{5} \overline{v}_1

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