Question

In the Balmer series of atomic spectra of hydrogen, there is a line corresponding to wavelength of 4344 A${}^o$. Calculate the number of higher orbits from which the electron drops to generate other line (R $\times$ c = 3.289 $\times$ 10${}^{15}$).

Answer 1

$v=\frac{c}{\lambda }$, $\lambda$ $=4344A^o=4344$ $\times$ 10${}^{-10}$ m

$v=\frac{3\times {10}^{8}m{s}^{-1}}{4344\times {10}^{-10}m}$ $=6.907\times$ 10${}^{14}$ s${}^{-1}$

In the Balmer series, electrons drops to n${}_2$ to n${}_1$ = 2. Let n${}_2$ is the energy levels from where e${}^{-}$ drops to n = 2
$v=Rc$[$\frac{1}{n_1^2}$ $-\frac{1}{n_2^2}$]
6.907 $\times$ 10${}^{14}$ = 3.289 $\times$ 10${}^{15}$[$\frac{1}{2^2}$ - $\frac{1}{n_2^2}$]

$\frac{6.907\times {10}^{14}}{3.289\times {10}^{15}}$ = [$\frac{1}{4}$ - $\frac{1}{n_2^2}$] = $\frac{n_2^2-4}{4n_2^2}$

4n${}^2$ $\times$ 6.907 = (n${}^2$ - 4) $\times$ 3.289 $\times$ 10

4n${}^2$ $\times$ 6.907 = 32.89n${}_2^2$ - 4 $\times$ 32.89

5.262n${}_2^2$ = 131.56,

n${}_2^2$ = $\frac{131.56}{5.262}$

n${}_2$ = 5.

Thus, the higher orbit from which the electron drops is 5.