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What electron transition in a hydrogen atom,starting from the orbit $n=7$ will produce infrared light of wavelength 2170nm?( $R_H=1.09677\times 10^7m^{-1})$

$\begin{array}{1 1}(a)\;n=2\;to\;n=3&(b)\;n=6\;to\;n=7\\(c)\;n=7\;to\;n=4&(d)\;n=4\;to\;n=7\end{array}$

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Electron transition from $n_2$ level is associated with wavelength $\lambda$ given by
$\large\frac{1}{\lambda}=$$R_H\big[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]$
$\lambda=2170nm=2170\times 10^{-9}m$
$R_H=1.09677\times 10^7m^{-1}$
$n_1=?n_2=7$
$\therefore \large\frac{10^9}{2170}=$$1.09677\times 10^7\big[\large\frac{1}{n_1^2}-\frac{1}{7^2}\big]$
$\big[\large\frac{1}{n_1^2}-\frac{1}{7^2}=$$0.0420$
$\large\frac{1}{n_1^2}$$=0.420+\large\frac{1}{49}$$=0.064$
$\therefore n_1=4$
Electron transition from $n=7$ to $n=4$ will produce infrared light of wavelength 2170nm
Hence (c) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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