Question

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

• A. 5
• B. 7
• C. 6
• D. 8

Answer 1

The correct option is A 5

v = $\frac{c}{\lambda }$, $\lambda$ = 4344 A = 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, e${}^{-}$ drops to n = 2 i.e n${}_1$ = 2. Let n${}_2$ is the energy levels from where e${}^{-}$ drops to n = 2v = Rc[$\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$]6.907 $\times$ 10${}^{14}$ = 3.289 $\times$ 10${}^{15}$[$\frac{1}{1^2}$ - $\frac{1}{n_2^2}$]
$\frac{6.907\times {10}^{14}}{3.289\times {10}^{15}}$ = [$\frac{1}{4}$ - $\frac{1}{n_2^2}$] = $\frac{n_2^2-4}{4n_2^2}$
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