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The ratio of maximum to minimum wavelength in Balmer series of an hydrogenic atom is
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The correct answer is:
$\frac{9}{5}$
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
$\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5}$
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
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
$\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5}$
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