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Calculate the wave length of light radiation that would be emitted,when an electron in the fourth Bohr's orbit of $He^+$ ion falls to the second Bohr's orbit.To what transition does this light radiation correspond in the H-atom?

$(a)\;2\qquad(b)\;1\qquad(c)\;3\qquad(d)\;4$

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Let us calculate $\Delta E$ first
$\Delta E=21.7\times 10^{-19}z^2(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Substituting $n_1=2$ and $n_2=4$ $z=2$
$\Delta E=1.632\times 10^{-18}J$
Now $\lambda=\large\frac{hC}{\Delta E}$
$\Rightarrow 1.218\times 10^{-17}m$
$\Rightarrow 1218A^{\large\circ}$
It means H-atom this transition would lie in Lyman series.Hence our aim is how to find the transition $m\to 1$
$\Rightarrow \Delta E=21.7\times 10^{-19}\times 1^2(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\Rightarrow 1.632\times 10^{-18}=21.7\times 10^{-19}(1-\large\frac{1}{n_2^2})$
$\Rightarrow n_2=2$
Hence the corresponding transition in H-atom is $2\to 1$
Hence (a) is the correct answer.
answered Jan 9, 2014 by sreemathi.v
 

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