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Calculate the wavelength of radiations emitted producing a line in Lyman series,when an electron falls from fourth stationary state in hydrogen atom $(R_H=1.1\times 10^7m^{-1})$

$\begin{array}{1 1}(a)\;1.34\times 10^{-7}m&(b)\;0.9696\times 10^{-7}m\\(c)\;1.34\times 10^{-4}m&(d)\;1.1\times 10^{-2}m\end{array}$

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1 Answer

$\large\frac{1}{\lambda}$$=R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$R_H=1.1\times 10^7$ for Lyman series $n_1=1$ and $n_2=4$ (Given)
$\therefore \large\frac{1}{\lambda}=$$1.1\times 10^7\bigg[\large\frac{1}{1^2}-\frac{1}{4^2}\bigg]$
$\lambda=0.9696\times 10^{-7}m$
Hence (b) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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