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A stationary $He^+$ ion emitted a photon corresponding to the first line $(H_{\alpha})$ of the Lyman series .That photon liberated a photo electron from a stationary H atom in ground state.What is the velocity of photo electron?$R_H=109678cm^{-1}$

$\begin{array}{1 1}(a)\;2.03\times 10^8cm/s&(b)\;3.09\times 10^8cm/s\\(c)\;3.09\times 10^2cm/s&(d)\;1.03\times 10^8cm/s\end{array}$

1 Answer

Energy of photon liberated from $He^+$
$H_{\alpha}$ line of Lyman series =$hcR_HZ^2\big[\large\frac{1}{1^2}-\frac{1}{2^2}\big]$
$\Rightarrow 6.625\times 10^{-27}\times 3\times 3\times 10^{10}\times 109678\times 2^2\big[\large\frac{3}{4}\big]$
$\Rightarrow 6.54\times 10^{-11}erg$
$6.54\times 10^{-11}=E_1$ of H +$\large\frac{1}{2}$$mu^2$
$\Rightarrow 13.6\times 1.602\times 10^{-21}+\large\frac{1}{2}$$mu^2$
$\large\frac{1}{2}$$mu^2=6.54\times 10^{-11}-2.179\times 10^{-11}$
$\Rightarrow 4.361\times 10^{-11}erg$
$u^2=\large\frac{4.361\times 10^{-11}\times 2}{9.108\times 10^{-28}}$
$u=3.09\times 10^8cm/s$
Hence (b) is the correct answer.
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