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The wavelength of the radiations emitted,when in a hydrogen atom electron falls from infinity to stationary state is ($R_H=1.097\times 10^7m^{-1})$

$\begin{array}{1 1}(a)\;9.1\times 10^{-8}nm\\(b)\;192nm\\(c)\;406nm\\(d)\;91nm\end{array}$

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$\large\frac{1}{\lambda}=$$R_H\big[\large\frac{1}{\infty}-\frac{1}{1^2}\big]$
$\therefore \lambda=\large\frac{1}{R_H}$
$\Rightarrow \large\frac{1}{1.097\times 10^7}$m
$\Rightarrow 9.1\times 10^{-8}m$
$\Rightarrow 91\times 10^{-9}m$
$\Rightarrow 91nm$
Hence (d) is the correct answer.
answered Jan 13, 2014 by sreemathi.v
 

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