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Calculate the shortest and longest wavelength in H spectrum of Lyman series. $R_H = 109678 cm^{-1}$

$(a)\;911.7A^{\large\circ},1215.67A^{\large\circ}\qquad(b)\;9117A^{\large\circ},1215.67A^{\large\circ}\qquad(c)\;91.17A^{\large\circ},121567A^{\large\circ}\qquad(d)\;911.7A^{\large\circ},121.67A^{\large\circ}$

1 Answer

For Lyman series $n_1=1$
For shortest $\lambda$ of Lyman series; energy difference in two levels showing transition should be maximum i.e $n_2 = \infty$
$\large\frac{1}{\lambda}=R_H[\large\frac{1}{1^2}-\large\frac{1}{\infty^2}]$
$\large\frac{1}{\lambda}=109678$
$\lambda = 911.7\times10^{-8} cm$
$=911.7A^{\large\circ}$
For longest $\lambda$ of Lyman series; energy difference in two levels showing transition should be minimum i.e $n_2 = 2$
$\large\frac{1}{\lambda}=R_H[\large\frac{1}{1^2}-\large\frac{1}{2^2}]$
$=109678\times\large\frac{3}{4}$
$\lambda = 1215.67\times10^{-8}cm$
$=1215.67A^{\large\circ}$
Hence answer is (a)
answered Feb 7, 2014 by sharmaaparna1
 

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