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Calculate the frequency of the spectral line emitted when the electron in n=3 in H-atom de-excites to ground state.$(R_H=109737/cm)$

$\begin{array}{1 1}(a)\;2.92\times 10^{15}/sec&(b)\;1.32\times 10^{16}/sec\\(c)\;3.4\times 10^{17}/sec&(d)\;1.32\times 10^{15}/sec\end{array}$

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$\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$\large\frac{c}{\lambda}$$=V=R_H.c\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$\Rightarrow 109737\times 3\times 10^{10}\bigg[\large\frac{1}{1^2}-\frac{1}{3^2}\bigg]$
$\Rightarrow 2.92\times 10^{15}/sec$
Hence (a) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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