JEE Main 2022 — Physics Modern Physics

2022integermedium

In a hydrogen spectrum $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be " $a\lambda$ " between the wavelength of ${3}^{\mathrm{rd}}$ transition line of Paschen series and that of ${2}^{\mathrm{nd}}$ transition line of Balmer Series where $a=$ _____.

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

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Solution

Using Rydberg formula,

$\frac{1}{\lambda }=R(\frac{1}{{n}_{f}^{2}}-\frac{1}{{n}_{i}^{2}})$ , where, $R$ is Rydberg constant and $n$ is quantum number.

For first line of Lyman series

$\frac{1}{\lambda }=R(1-\frac{1}{4})=R(\frac{3}{4})$

$\Rightarrow \lambda =\frac{4}{3R}...(1)$

For ${3}^{\mathrm{rd}}$ line of Paschen series

$\frac{1}{{\lambda }_{3}}=R(\frac{1}{{3}^{2}}-\frac{1}{{6}^{2}})=\frac{R}{9}\times \frac{3}{4}$

For ${2}^{\mathrm{nd}}$ line of Balmer series

$\frac{1}{{\lambda }_{2}}=R(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}})=\frac{R}{4}\times \frac{3}{4}$

Thus, $a\lambda ={\lambda }_{3}-{\lambda }_{2}=\frac{12}{R}-\frac{16}{3R}=\frac{20}{3R}$

Putting equation $(1)$

$a(\frac{4}{3R})=\frac{20}{3R}\Rightarrow a=5$

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JEE Main 2022 — Physics Modern Physics

2022integermedium

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

Using Rydberg formula,

$\frac{1}{\lambda }=R(\frac{1}{{n}_{f}^{2}}-\frac{1}{{n}_{i}^{2}})$ , where, $R$ is Rydberg constant and $n$ is quantum number.

For first line of Lyman series

$\frac{1}{\lambda }=R(1-\frac{1}{4})=R(\frac{3}{4})$

$\Rightarrow \lambda =\frac{4}{3R}...(1)$

For ${3}^{\mathrm{rd}}$ line of Paschen series

$\frac{1}{{\lambda }_{3}}=R(\frac{1}{{3}^{2}}-\frac{1}{{6}^{2}})=\frac{R}{9}\times \frac{3}{4}$

For ${2}^{\mathrm{nd}}$ line of Balmer series

$\frac{1}{{\lambda }_{2}}=R(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}})=\frac{R}{4}\times \frac{3}{4}$

Thus, $a\lambda ={\lambda }_{3}-{\lambda }_{2}=\frac{12}{R}-\frac{16}{3R}=\frac{20}{3R}$

Putting equation $(1)$

$a(\frac{4}{3R})=\frac{20}{3R}\Rightarrow a=5$

Your note

JEE Main Modern Physics in other years

Keep practicing Modern Physics

Work through every JEE Main Modern Physics PYQ, year by year.

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JEE Main 2022 — Physics Modern Physics

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2022 — Physics Modern Physics

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics