JEE Main 2023 — Chemistry Physical Chemistry

2023mcqeasy

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n=4$ to $n=2$ of ${\mathrm{He}}^{+}$ spectrum

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Options

A
$n=2$ to $n=1$
B
$n=1$ to $n=3$
C
$n=1$ to $n=2$
D
$n=3$ to $n=4$

Did you get this right?

Solution

For ${\mathrm{He}}^{+}$ ion, the wave number associated with the Balmer transition, $n=4$ to $n=2$ is given by:

$\frac{1}{\lambda }={\mathrm{RZ}}^{2}(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}})$

Where, ${n}_{1}=2$

${n}_{2}=4$

$Z$ = atomic number of helium

$\frac{1}{\lambda }=R(2{)}^{2}[\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}]$

$\Rightarrow \frac{1}{\lambda }=4R[\frac{4-1}{16}]$

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

$\Rightarrow \lambda =\frac{4}{3R}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of ${\mathrm{He}}^{+}$ .

$\lambda (H)=\lambda ({\mathrm{He}}^{+})$

$R(1{)}^{2}[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}]=\frac{3R}{4}$

By hit and trial method,

$\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}=\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}$

On comparing ${n}_{1}=1&{n}_{2}=2$

So, the correct answer is n = 2 to n = 1.

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

2023mcqeasy

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n=4$ to $n=2$ of ${\mathrm{He}}^{+}$ spectrum

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Options

A
$n=2$ to $n=1$
B
$n=1$ to $n=3$
C
$n=1$ to $n=2$
D
$n=3$ to $n=4$

Did you get this right?

Solution

For ${\mathrm{He}}^{+}$ ion, the wave number associated with the Balmer transition, $n=4$ to $n=2$ is given by:

$\frac{1}{\lambda }={\mathrm{RZ}}^{2}(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}})$

Where, ${n}_{1}=2$

${n}_{2}=4$

$Z$ = atomic number of helium

$\frac{1}{\lambda }=R(2{)}^{2}[\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}]$

$\Rightarrow \frac{1}{\lambda }=4R[\frac{4-1}{16}]$

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

$\Rightarrow \lambda =\frac{4}{3R}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of ${\mathrm{He}}^{+}$ .

$\lambda (H)=\lambda ({\mathrm{He}}^{+})$

$R(1{)}^{2}[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}]=\frac{3R}{4}$

By hit and trial method,

$\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}=\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}$

On comparing ${n}_{1}=1&{n}_{2}=2$

So, the correct answer is n = 2 to n = 1.

Your note

JEE Main Physical Chemistry in other years

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

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JEE Main 2023 — Chemistry Physical Chemistry

Options

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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

Options

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