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Find the de Broglie $ \lambda$ of a neutron, in thermal equilibrium with matter, having an average $KE\: of \bigg(\large\frac{3}{2} \bigg)\: kT$ at $300 \: K.$

$\begin {array} {1 1} (a)\;14.6 \times 10^{-10}m & \quad (b)\;0.146\: nm \\ (c)\;14.6\: nm & \quad (d)\;1.46 × 10^{-9}m \end {array}$

 

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Ans : (b)
$T=300\: K\: \: \: \: \: \: k=1.38 \times 10^{-23}kg m^2 / s^2K$
Average $KE,\: K’ = \large\frac{3}{2} kT$
$= \large\frac{3}{2} \times 1.38 \times 106{-23} \times 300 = 6.21 \times 10^{-21} J$
de Broglie wavelength, $ \lambda’ = \large\frac{h}{ \sqrt {(2K’mn)}}$
$= \large\frac{6.63 \times 10^{-34}}{\sqrt{(2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27 }) }}$
$= 1.46 \times 10^{-10}m = 0.146\: nm$

 

answered Dec 27, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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