JEE Main 2026 Apr 6 Shift 1, Chemistry Q2: Bohr's Atomic Model

JEE Main2026Apr 6, Shift 1Chemistry

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

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

Solution

Apply the Rydberg formula $\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ .

H, Lyman shortest ( $n_{1} = 1$ , $n_{2} \to \infty$ ): $\frac{1}{x} = R (1)^{2} (1) = R$ .

$He^{+}$ , Balmer longest ( $n_{1} = 2$ , $n_{2} = 3$ , $Z = 2$ ): $\frac{1}{λ} = R (4) (\frac{1}{4} - \frac{1}{9}) = 4 R \cdot \frac{5}{36} = \frac{5 R}{9}$ .

Divide: $\frac{x}{λ} = \frac{5}{9} \Rightarrow λ = \frac{9 x}{5}$ .

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JEE Main 2026 Apr 6 Shift 1, Chemistry Q2: Bohr's Atomic Model

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