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The wavelength of $\beta$ line of the Balmer Series is $4815 A^{\large\circ}$. What is the wavelength of $\alpha$ line of balmer Series of the same atom?

$(a)\;4500A^{\large\circ}\qquad(b)\;5600A^{\large\circ}\qquad(c)\;2600A^{\large\circ}\qquad(d)\;6500A^{\large\circ}$

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First Line is called $\alpha$ line
Second line is called as $\beta$ line
$\large\frac{1}{\lambda} = \overline R_HZ^2[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
For Balmer Series $n_1$=2 and for $\beta$ line $n_2$=4
$\large\frac{1}{\lambda_\beta} = \overline R_HZ^2[\large\frac{1}{2^2}-\large\frac{1}{4^2}]$
$=\large\frac{3}{16} \overline R_HZ^2$
For $\alpha$-line $n_2$ = 3 For same series
$\large\frac{1}{\lambda_\beta} = \overline R_HZ^2[\large\frac{1}{2^2}-\large\frac{1}{3^2}]$
$=\large\frac{5}{16} \overline R_HZ^2$
$\large\frac{\lambda_{\beta'}}{\lambda_\beta}=\large\frac{3}{16}\times \large\frac{36}{5}$
$\lambda_{\beta'} = 4815\times10^{-10}\times \large\frac{108}{80}$
$=6.5\times10^{-7}$
$=6500A^{\large\circ}$
Hence answer is (d)
answered Feb 4, 2014 by sharmaaparna1
 

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