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Energy of the electron in H-atom is 1.50 times as much as energy as the minimum required for its escape (13.6eV) from the atom.Calculate the wavelength of the emitted electron.

$\begin{array}{1 1}(a)\;4.71A^{\large\circ}&(b)\;3.72A^{\large\circ}\\(c)\;4A^{\large\circ}&(d)\;5.6A^{\large\circ}\end{array}$

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If $E_1=13.6eV$
$E_2=13.6\times 1.50eV=20.4eV$
$\Delta E=(E_2-E_1)=20.4-13.6=6.8eV$
$\Rightarrow 6.8\times 1.6\times 10^{-19}J$
$\lambda=\large\frac{h}{\sqrt{2m(KE)}}$
$\Rightarrow \large\frac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 6.8\times 1.6\times 10^{-19}}}$
$\Rightarrow 4.71\times 10^{-10}m$
$\Rightarrow 4.71A^{\large\circ}$
Hence (a) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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