The ratio of wavelength of spectral lines H_a | vedclass

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(A) The Rydberg formula for the wavelength of spectral lines in the $H$-atom is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^

Vedclass · Accepted Answer

$27$

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(A) The Rydberg formula for the wavelength of spectral lines in the $H$-atom is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Balmer series,$n_1 = 2$.For the $H_\alpha$ line,$n_2 = 3$. Thus,$\frac{1}{\lambda_{H_\alpha}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = \frac{5R}{36}$.For the $H_\beta$ line,$n_2 = 4$. Thus,$\frac{1}{\lambda_{H_\beta}} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{4} - \frac{1}{16} ight] = \frac{3R}{16}$.Taking the ratio of the wavelengths: $\frac{\lambda_{H_\alpha}}{\lambda_{H_\beta}} = \frac{\lambda_{H_\beta}^{-1}}{\lambda_{H_\alpha}^{-1}} = \frac{3R/16}{5R/36} = \frac{3}{16} 	imes \frac{36}{5} = \frac{3 	imes 9}{4 	imes 5} = \frac{27}{20}$.Comparing this with $\frac{x}{20}$,we get $x = 27$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

The ratio of wavelength of spectral lines $H_\alpha$ and $H_\beta$ in the Balmer series is $\frac{x}{20}$. The value of $x$ is $...............$

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