Q156 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
The ratio of maximum to minimum wavelength in Balmer series of an hydrogenic atom is
- Correct answerA$\frac{9}{5}$
- B$\frac{12}{7}$
- C$\frac{9}{7}$
- D$\frac{14}{9}$
Solution
Wavelength of spectral lines of Hydrogen like atom is given as
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For Balmer series, $n_1=2$ and $n_2=3,4,5,6, \ldots$.
For minimum wavelength, $n_2=\infty$ From Eq. (i), $\frac{1}{\lambda_{\min }}=R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)$
For maximum wavelength, $n_2=3$
$\therefore$ From Eq. (i), we have
$\therefore$ From Eq. (ii) and (iii), we have $\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5}$
For minimum wavelength, $n_2=\infty$ From Eq. (i), $\frac{1}{\lambda_{\min }}=R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)$
For maximum wavelength, $n_2=3$
$\therefore$ From Eq. (i), we have
$\therefore$ From Eq. (ii) and (iii), we have $\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5}$
Hence, the correct answer is option: 'A'.