Q60 / 75 · JEE Main 2026 (22 January Shift 2) · Physics
The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly $\_\_\_\_$ nm .
- A1550
- Correct answerB1217
- C1875
- D1784
Solution
Smallest wavelength of lyman
$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right)$
$R = \dfrac{1}{\lambda} = \dfrac{1}{91} \text{ nm}^{-1}$
$\lambda_{max}$ for balmer series
$n_1 = 2 \rightarrow n_2 = 3$
$\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$
$\dfrac{1}{\lambda_B} = \dfrac{1}{91}\left(\dfrac{5}{36}\right)$
$\lambda_B = \left(\dfrac{91 \times 36}{5}\right) = 655.2 \text{ nm}$
$\lambda_{max}$ paschen
$n_1 = 3 \rightarrow n_2 = 4$
$\dfrac{1}{\lambda_p} = \dfrac{1}{91}\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right) = \dfrac{1}{91} \times \dfrac{7}{144}$
$\lambda_p = \left(\dfrac{91 \times 144}{7}\right) = 1872 \text{ nm}$
$\Delta\lambda = \lambda_P - \lambda_B = 1872 - 655.2$
$\Delta\lambda = 1216.8$
$\Delta\lambda \approx 1217$
$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right)$
$R = \dfrac{1}{\lambda} = \dfrac{1}{91} \text{ nm}^{-1}$
$\lambda_{max}$ for balmer series
$n_1 = 2 \rightarrow n_2 = 3$
$\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$
$\dfrac{1}{\lambda_B} = \dfrac{1}{91}\left(\dfrac{5}{36}\right)$
$\lambda_B = \left(\dfrac{91 \times 36}{5}\right) = 655.2 \text{ nm}$
$\lambda_{max}$ paschen
$n_1 = 3 \rightarrow n_2 = 4$
$\dfrac{1}{\lambda_p} = \dfrac{1}{91}\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right) = \dfrac{1}{91} \times \dfrac{7}{144}$
$\lambda_p = \left(\dfrac{91 \times 144}{7}\right) = 1872 \text{ nm}$
$\Delta\lambda = \lambda_P - \lambda_B = 1872 - 655.2$
$\Delta\lambda = 1216.8$
$\Delta\lambda \approx 1217$
Hence, the correct answer is option: 'B'.