Question

The wavelength of the first member of the Balmer series in the hydrogen spectrum is $6563 Å$. What is the wavelength of the first member of Lyman series?

Answer 1

SOLUTION : For Balmer series, first spectral line,

$\frac{1}{\lambda_{1}}=R_{H}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) ; \quad \lambda_{1}=\frac{36}{5 R_{H}}$

For Lyman series, first spectral line $\frac{1}{\lambda_{2}}=R_{H}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right), \lambda_{2}=\frac{4}{3 R_{H}}$

From equations (i) and (ii) $\frac{\lambda_{2}}{\lambda_{1}}=\frac{4}{3 R_{H}} \times \frac{5 R_{H}}{36}=\frac{5}{27}, \quad \lambda_{2}=\frac{5}{27} \times \lambda_{1}$

Given, $\lambda_{1}=6563 Å, \lambda_{2}=? \quad \lambda_{2}=\frac{5}{27} \times 6563 Å=1215 Å$.