Physics - Modern Physics Question with Solution | TestHub

For Balmer Series:

Shortest wavelength, $\frac{E_o}{4}=\frac{hc}{{\lambda }_s}$

Longest wavelength, $\left(\frac{E_o}{4}-\frac{E_0}{9}\right)=\frac{hc}{{\lambda }_l}=\frac{5E_o}{36}$

$\therefore \frac{{\lambda }_l}{{\lambda }_s}=\frac{9}{5}$

Range of energy level in Balmer series,

$\left(\frac{5E_0}{36}, \frac{E_0}{4}\right)$

Range of energy level in Paschen series,

$\left(\frac{7E_0}{144}, \frac{E_0}{9}\right)$

Hence, no overlap.

For Lyman series:

For shortest wavelength,

$E_0=\frac{hc}{{\lambda }_0}$

For general wavelength,

$\left(E_0-\frac{E_0}{n^2}\right)=\frac{hc}{\lambda }$

$\Rightarrow \frac{1}{\left(1-\frac{1}{n^2}\right)}=\frac{\lambda }{{\lambda }_0}$

$\therefore \lambda =\left(\frac{n^2}{n^2-1}\right){\lambda }_0$

or $\lambda =\left(1+\frac{1}{n^2-1}\right){\lambda }_0$$=\left(1+\frac{1}{m^2}\right){\lambda }_0$

as $m$ should be an integer.

Range of energy level in Lyman series, $\left(\frac{3E_0}{4}, E_0\right)$

$\Rightarrow$ Lyman and Balmer series wavelengths do not overlap.