Question

The wavelength of the second line of the balmer series in the hydrogen spectrum is 4861 A calculate the wavelength of the first line

Answer 1

Answer:

6562 A⁰

Explanation:

Balmer series is the spectral series emitted when electron jumps from a higher orbital to $2^{nd}$ orbital of unipositive hydrogen like-species.

$2^{nd}$ line indicates transition from 4 --> 2

$1^{st}$ line indicates transition from 3 -->2

let λ be represented by L

Using the following relation for wavelength;

$\boxed{hc(\frac{1}{L}) = 13.6z^2(\frac{1}{n^2}-\frac{1}{n'^2})}$

$L=\frac{hc(n^2)(n'^2)}{13.6(z^2)(n'^2-n^2)}$

$L=k\frac{(n^2)(n'^2)}{(n'^2-n^2)}$

For 4-->2 transition

L=4861 = $k\frac{(16)(4)}{12} =\frac{16}{3}k$

$\boxed{k=911.4}$

For 3-->2 transition

$L'=k\frac{36}{5} =911.4X7.2$=6562 A⁰

Hope this answer helped you :)