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:
$\frac{\lambda_2}{656} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{20}{27}$
$\lambda_2 = 656 	imes \frac{20}{27} \approx 485.9 \approx 486 	ext{ nm}$

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