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An electron moves in an electric field with a kinetic energy of 2.5eV.What is the associated De-Broglie wavelength?

$\begin{array}{1 1}(a)\;6.7\times 10^{-8}cm&(b)\;7.7\times 10^{-8}cm\\(c)\;7.7\times 10^{-2}cm&(d)\;6.5\times 10^{-8}cm\end{array}$

1 Answer

K.E=$\large\frac{1}{2}$$mu^2$
$\Rightarrow \large\frac{1}{2}$$m\big[\large\frac{h}{m\lambda}\big]^2$
K.E=$\large\frac{1}{2}\frac{h^2}{m\lambda^2}$
$\therefore \lambda^2=\large\frac{h^2}{2m(K.E)}$
$\lambda=\sqrt{\large\frac{h^2}{2mK.E}}$
$\Rightarrow \large\frac{6.627\times 10^{-27}}{\sqrt{2\times 9.108\times 10^{-28}\times 2.5\times 1.602\times 10^{-12}}}$
$\Rightarrow 7.7\times 10^{-8}$cm
Hence (b) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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