Bohr's Model

Question

The wavelength of the first line in the Balmer series of hydrogen spectrum is 656 nm. The wavelength of the second line of this series is approximately:

RankGuru · Accepted Answer

The wavelength of spectral lines is given by the **Rydberg Formula**:
$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$
For the **Balmer Series**, $n_1 = 2$.
1. **First Line ($H_\alpha$ line)**: Transition from $n_2 = 3 	o n_1 = 2$.
$\frac{1}{656} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = R_H \left( \frac{5}{36} ight)$
2. **Second Line ($H_\beta$ line)**: Transition from $n_2 = 4 	o n_1 = 2$.
$\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{16} ight) = R_H \left( \frac{3}{16} ight)$
Dividing Eq. 1 by Eq. 2:
$\frac{\lambda_2}{656} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{20}{27}$
$\lambda_2 = 656 	imes \frac{20}{27} \approx 485.9 \approx 486 	ext{ nm}$

Rank Guru

Bohr's Model

Select the correct option:

Solution

🔒 Solution Hidden from View

More bohr's model Practice Questions

The ionisation energy of a hydrogen-like ion He⁺ in its ground state is (ionisation energy of H = 13...

The radius of the third orbit of hydrogen atom according to Bohr's model is:

According to Bohr's model, the energy of an electron in the nth orbit of a hydrogen atom is given by...

About This Question

Solution