Question

The first three spectral lines of $\mathrm{H}$-atom in the Balmer series are given${\mathrm{\lambda }}_1$,${\mathrm{\lambda }}_2$, ${\mathrm{\lambda }}_3$and considering the Bohr atomic model, the wavelengths of the first and third spectral lines $({\mathrm{\lambda }}_1/{\mathrm{\lambda }}_3)$are related by a factor of approximately$\text{'}\mathrm{x}\text{'}\times {10}^{-1}$. The value of $\mathrm{x}$, to the nearest integer, is ________.

Answer 1

The value of$\mathrm{x}$, the nearest integer, is$15$.

• The Explanation: The first three spectral lines of $\mathrm{H}$-atom in the Balmer series are given ${\mathrm{\lambda }}_1$${\mathrm{\lambda }}_2$${\mathrm{\lambda }}_3$.

For first-line is

$\frac{1}{{\mathrm{\lambda }}_1}={\mathrm{Rz}}^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$

$\frac{1}{{\mathrm{\lambda }}_1}={\mathrm{Rz}}^2\frac{5}{36}$------(I)

The third line is

$\frac{1}{{\mathrm{\lambda }}_3}={\mathrm{Rz}}^2\left[\frac{1}{2^2}-\frac{1}{5^2}\right]$

$\frac{1}{{\mathrm{\lambda }}_3}={\mathrm{Rz}}^2\frac{21}{100}$-------(II)

$\left(\mathrm{II}\right)÷\left(\mathrm{I}\right)$

$\frac{{\lambda }_1}{{\mathrm{\lambda }}_3}=\frac{21}{100}\times \frac{36}{5}$

$=15.12\times {10}^{-1}$

$\mathrm{x}=15$

Hence, the value of $\mathrm{x}$, the nearest integer, is$15$