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If the KE of a free electron doubles, its de-Broglie wavelength changes by the factor

$\begin {array} {1 1} (a)\;\large\frac{1}{\sqrt 2} & \quad (b)\;\sqrt 2 \\ (c)\;\large\frac{1}{2} & \quad (d)\;2 \end {array}$

 

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Ans : (a)
$ \lambda=\large\frac{h}{mv} = \large\frac{h}{p} = \large\frac{h}{\sqrt {2mE_k}}$
$ \lambda \sim \large\frac{1}{\sqrt{E_k}}$
$ \lambda’ \sim \large\frac{1}{\sqrt{2E_k}}$

 

answered Dec 28, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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