Question

If wavelength of the first line of the Paschen series of hydrogen atom is _____ nm, then the wavelength of the second line of this series is nm (Nearest integer)

Answer 1

Correct Answer: 492

The Paschen series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (\( n_2 \)) to the third energy level (\( n_1 = 3 \)).
The wavelength of emitted radiation is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \)) - \( n_1 = 3 \) (for the Paschen series) - \( n_2 \) is the upper energy level (\( n_2 = 4 \) for the first line and \( n_2 = 5 \) for the second line)
Step 1: Calculate the wavelength of the first line (\( n_2 = 4 \)). Given \( \lambda = 720 \, \text{nm} \) for the first line: \[ \frac{1}{720 \times 10^{-9}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{720 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right) \] \[ \frac{1}{720 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{16 - 9}{144} \right) \] \[ \frac{1}{720 \times 10^{-9}} = 1.097 \times 10^7 \times \frac{7}{144} \] This confirms the consistency of the Rydberg formula for the given wavelength.
Step 2: Calculate the wavelength of the second line (\( n_2 = 5 \)). For \( n_2 = 5 \): \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{25} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{25 - 9}{225} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{16}{225} \] \[ \frac{1}{\lambda} = 7.809 \times 10^5 \, \text{m}^{-1} \] \[ \lambda = \frac{1}{7.809 \times 10^5} = 4.92 \times 10^{-7} \, \text{m} = 492 \, \text{nm} \]