JEE Main 2023 — Chemistry Physical Chemistry

2023integereasy

If wavelength of the first line of the Paschen series of hydrogen atom is $720\mathrm{nm}$ , then the wavelength of the second line of this series is _______ $\mathrm{nm}$ . (Nearest integer)

Official previous-year question

Held on 24 Jan 2023 · Verified 6 Jul 2026.

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Solution

Paschen series are spectral lines coming to $n=3$

${1}^{\mathrm{st}}$ line of Paschen series makes a transition from, $4\rightarrow 3$

So, ${2}^{\mathrm{nd}}$ line of Paschen series makes a transition from, $5\rightarrow 3$

We know that,

$\frac{1}{\lambda }=R[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}]{Z}^{2}$

For ${1}^{\mathrm{st}}$ line, $\frac{1}{\lambda }=R[\frac{1}{(3{)}^{2}}-\frac{1}{(4{)}^{2}}]{Z}^{2}$

and for ${2}^{\mathrm{nd}}$ line,

$\frac{1}{{\lambda }^{'}}=R[\frac{1}{(3{)}^{2}}-\frac{1}{(5{)}^{2}}]{Z}^{2}$

or $\frac{(\frac{1}{\lambda })}{(\frac{1}{{\lambda }^{'}})}=\frac{R[\frac{7}{9\times 16}]{Z}^{2}}{R[\frac{16}{9\times 25}]{Z}^{2}}$

Given, $\lambda =720\mathrm{nm}$

or $\frac{(\frac{1}{720})}{(\frac{1}{\lambda '})}=\frac{175}{256}$

or $\frac{{\lambda }^{'}}{720}=\frac{175}{256}$

or ${\lambda }^{'}\approx 492\mathrm{nm}$

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JEE Main 2023 — Chemistry Physical Chemistry

2023integereasy

If wavelength of the first line of the Paschen series of hydrogen atom is $720\mathrm{nm}$ , then the wavelength of the second line of this series is _______ $\mathrm{nm}$ . (Nearest integer)

Official previous-year question

Held on 24 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Paschen series are spectral lines coming to $n=3$

${1}^{\mathrm{st}}$ line of Paschen series makes a transition from, $4\rightarrow 3$

So, ${2}^{\mathrm{nd}}$ line of Paschen series makes a transition from, $5\rightarrow 3$

We know that,

$\frac{1}{\lambda }=R[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}]{Z}^{2}$

For ${1}^{\mathrm{st}}$ line, $\frac{1}{\lambda }=R[\frac{1}{(3{)}^{2}}-\frac{1}{(4{)}^{2}}]{Z}^{2}$

and for ${2}^{\mathrm{nd}}$ line,

$\frac{1}{{\lambda }^{'}}=R[\frac{1}{(3{)}^{2}}-\frac{1}{(5{)}^{2}}]{Z}^{2}$

or $\frac{(\frac{1}{\lambda })}{(\frac{1}{{\lambda }^{'}})}=\frac{R[\frac{7}{9\times 16}]{Z}^{2}}{R[\frac{16}{9\times 25}]{Z}^{2}}$

Given, $\lambda =720\mathrm{nm}$

or $\frac{(\frac{1}{720})}{(\frac{1}{\lambda '})}=\frac{175}{256}$

or $\frac{{\lambda }^{'}}{720}=\frac{175}{256}$

or ${\lambda }^{'}\approx 492\mathrm{nm}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry