Question

The Balmer series consists of the spectral lines from hydrogen for an electron making a transition from an excited state to the m = 2 state. The Lyman series consists of the spectral lines from hydrogen for an electron making a transition from an excited state to the m = 1 state. Determine the wavelengths of the first four spectral lines of the Lyman series (n = 2, 3, 4, and 5)

Answer 1

Answer:

Wavelength of the first four spectral line:

$\lambda_1=1216{\AA}$

$\lambda_2=1026{\AA}$

$\lambda_3=972.8{\AA}$

$\lambda_4=950{\AA}$

Explanation:

Lman series when electron from higher shell number and jumps to 1st shell then all the possible lines is called as lyman series.

to calculate wavelength of first four spectral line:

For hydrogen Z=1;

by using rydberg equation

$\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]$

1. n=2 to n=1

$\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]$

$\frac{1}{R} =912{\AA}$ =rydberg constant

$\lambda=\frac{4}{3R}$

$\lambda_1=1216{\AA}$

2. n=3 to n=1

$\lambda=\frac{9}{8R}$

$\lambda_2=1026{\AA}$

3. n=4 to n=1

$\lambda=\frac{16}{15R}$

$\lambda_3=972.8{\AA}$

4. n=5 to n=1

$\lambda=\frac{25}{24R}$

$\lambda_4=950{\AA}$