Q1 / 75 · JEE Main 2026 (24 January Shift 2) · Chemistry
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
- A$1.14 \times 10^{-7} \mathrm{~cm}$
- Correct answerB$1.14 \times 10^{-6} \mathrm{~cm}$
- C$2.28 \times 10^{-7} \mathrm{~cm}$
- D$2.28 \times 10^{-6} \mathrm{~cm}$
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
$\lambda = 1.14 \times 10^{-6}$ cm
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
$\lambda = 1.14 \times 10^{-6}$ cm
Hence, the correct answer is option: 'B'.