The wavelength of spectral line obtained in the spectrum of Li^ 2+ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

Correct answer: B. 1.14 × 10^ -6 ~cm For Li²⁺ (Z=3): From n_2 + n_1 = 4 and n_2 - n_1 = 2 , we get n_2 = 3 , n_1 = 1 Using Bohr wavelength formula: (1)/(λ) = RZ^2((1)/(n_1^2) - (1)/(n_2^2)) (1)/(λ) = 1.097 × 10^7 × 9 × (1 - (1)/(9)) = 9.873 × 10^7 × (8)/(9) (1)/(λ) = 8.776 × 10^7 m⁻¹ λ = 1.14 × 10^ -6 cm

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Structure of Atom

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Structure of Atom

The wavelength of spectral line obtained in the spectrum of Li^ 2+...

The wavelength of spectral line obtained in the spectrum of

\mathrm{Li}^{2+}

ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

Options

A $1.14 \times 10^{-7} \mathrm{~cm}$
B $1.14 \times 10^{-6} \mathrm{~cm}$
C $2.28 \times 10^{-7} \mathrm{~cm}$
D $2.28 \times 10^{-6} \mathrm{~cm}$

Correct answer

B. $1.14 \times 10^{-6} \mathrm{~cm}$

Step-by-step solution

For Li²⁺ (Z=3): From

n_2 + n_1 = 4

and

n_2 - n_1 = 2

, we get

n_2 = 3

n_1 = 1

Using Bohr wavelength formula:

\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

\frac{1}{\lambda} = 1.097 \times 10^7 \times 9 \times \left(1 - \frac{1}{9}\right) = 9.873 \times 10^7 \times \frac{8}{9}

\frac{1}{\lambda} = 8.776 \times 10^7

m⁻¹

\lambda = 1.14 \times 10^{-6}

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