JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

If shortest wavelength of hydrogen atom in Lyman series is $x$ , then longest wavelength in Balmer series of $\text{He}^+$ is:

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{9x}{5}$
B
$\dfrac{36x}{5}$
C
$\dfrac{x}{4}$
D
$\dfrac{5x}{9}$

Did you get this right?

Solution

For the shortest wavelength in the Lyman series of the hydrogen atom ( $Z=1$ ), the transition is from $n_2 = \infty$ to $n_1 = 1$ .

$\dfrac{1}{x} = R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R$

This gives $R = \dfrac{1}{x}$ .

For the longest wavelength in the Balmer series of $\text{He}^+$ ( $Z=2$ ), the transition is from $n_2 = 3$ to $n_1 = 2$ .

$\dfrac{1}{\lambda} = R(2)^2 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)$

$\dfrac{1}{\lambda} = 4R \left(\dfrac{1}{4} - \dfrac{1}{9}\right)$

$\dfrac{1}{\lambda} = 4R \left(\dfrac{5}{36}\right) = \dfrac{5R}{9}$

Substituting $R = \dfrac{1}{x}$ :

$\dfrac{1}{\lambda} = \dfrac{5}{9x}$

$\lambda = \dfrac{9x}{5}$

Answer: $\dfrac{9x}{5}$

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

If shortest wavelength of hydrogen atom in Lyman series is $x$ , then longest wavelength in Balmer series of $\text{He}^+$ is:

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{9x}{5}$
B
$\dfrac{36x}{5}$
C
$\dfrac{x}{4}$
D
$\dfrac{5x}{9}$

Did you get this right?

Solution

For the shortest wavelength in the Lyman series of the hydrogen atom ( $Z=1$ ), the transition is from $n_2 = \infty$ to $n_1 = 1$ .

$\dfrac{1}{x} = R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R$

This gives $R = \dfrac{1}{x}$ .

For the longest wavelength in the Balmer series of $\text{He}^+$ ( $Z=2$ ), the transition is from $n_2 = 3$ to $n_1 = 2$ .

$\dfrac{1}{\lambda} = R(2)^2 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)$

$\dfrac{1}{\lambda} = 4R \left(\dfrac{1}{4} - \dfrac{1}{9}\right)$

$\dfrac{1}{\lambda} = 4R \left(\dfrac{5}{36}\right) = \dfrac{5R}{9}$

Substituting $R = \dfrac{1}{x}$ :

$\dfrac{1}{\lambda} = \dfrac{5}{9x}$

$\lambda = \dfrac{9x}{5}$

Answer: $\dfrac{9x}{5}$

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 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — 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