The first three spectral lines of H-atom in the Balmer series are given by $\lambda_1,\lambda_2,\lambda_3$ respectively. Considering the Bohr atomic model, the ratio of wave lengths of first and third spectral lines $\frac {\lambda_1}{\lambda_3}$ is approximately given by $’x’ \times 10^{-1}$.

The value of x, to the nearest integer is———–.

Solution

$\frac{1}{{{\lambda _1}}} \propto \frac{1}{4} – \frac{1}{9} = \frac{5}{{36}}$

$\frac{1}{{{\lambda _3}}} \propto \frac{1}{4} – \frac{1}{{25}} = \frac{{21}}{{100}}$

$\frac{{{\lambda _1}}}{{{\lambda _3}}} = \frac{{21}}{{100}} \times \frac{{36}}{5} = \frac{{21 \times 9}}{{25 \times 5}} = 15.12 \times {10^{ – 1}}$

$\therefore x = 15$