Question

The wavelength of $\beta$-line of the Balmer series is $4815A^{\circ }$. What is the wavelength of $\alpha$-line of the Balmer series of the same atom?

Answer 1

For balmer series,

$\alpha$-line $=n_1=2,n_2=3$

$\beta$-line$=n_1=2n_2=4$

Now $\frac{1}{\lambda }$$=109677$$(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$=109677(\frac{1}{4}-\frac{1}{9})$

=$\frac{109677\times 5}{4\times 9}$

$\frac{1}{\lambda }$ $=15232.917cm$${}^{-1}$

$\lambda =6.5647x{10}^{-5}cm$

$\lambda$$=6564.7A$${}^0$ = for $\alpha$ line of Balmer series