Question

The first member of the Balmer series of hydrogen atom has wavelength of $656.3nm$, Calculate the wavelength and frequency of the second member of the same series. Given : $C=3\times {10}^{8}m{s}^{"1}$.

Answer 1

We have
$\frac{1}{{\lambda }_0}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For first member of Balmer series
$n_1=2$ and $n_2=3$
${\lambda }_1=6563A^{\circ }$
$\frac{1}{6563\times {10}^{-10}}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}....\left(1\right)$
For second member of Balmer series, $n_1=2$ and $n_2=4$
$\frac{1}{{\lambda }_2}=R(\frac{1}{4}-\frac{1}{16})=\frac{3R}{16}....\left(2\right)$
Dividing equation $\left(1\right)$ by equation $\left(2\right)$
$\frac{{\lambda }_2}{6563\times {10}^{-10}}=\frac{5}{36}\times \frac{16}{3}$
${\lambda }_2=\frac{5\times 16\times 6563\times {10}^{-10}}{108}=4861A^{\circ }$
Frequence $v=\frac{C}{{\lambda }_2}=\frac{3\times {10}^8}{4861\times {10}^{-10}}=0.0006171\times {10}^{18}$
$v=6.17\times {10}^{14}Hz$.