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Calculate the wavelength of the first line and the series limit for the Lyman series for hydrogen.(Series limit is for the case when $n_2 = \infty)$

$(a)\;121.6nm\qquad(b)\;211.6nm\qquad(c)\;111.3nm\qquad(d)\;124.3nm$

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$\overline V = \large\frac{1}{\lambda} = \overline R_H Z^2[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
$\overline V_{max} = \large\frac{1}{\lambda_{min}}$
$=\large\frac{\overline R_H}{n_1^2}=\overline R_H = 10967700 m^{-1}$
$\lambda_{min}$ = 91.2 nm for the series limit
First line is obtained whwn $n_2 =2$ $n_1 = 1$
$\large\frac{1}{\lambda} = \overline R_H\times0.75$
$\lambda$ = 121.6nm
Hence the answer is (a)
answered Feb 5, 2014 by sharmaaparna1
 

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