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

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

 

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Ans : (c)
$ \lambda=\large\frac{h}{mv} \: \: and\: \: K =\large\frac{1}{2}\: mv^2 = \large\frac{(mv)^2} {2m}$
$ \Rightarrow mv = \sqrt{2mK}$
So, $ \lambda=\large\frac{h}{\sqrt{2mK}}$
$ \Rightarrow  \lambda \sim \large\frac{1}{\sqrt K}$
$ \large\frac{ \lambda_2}{\lambda_1}= \large\frac{ \sqrt{K_1}}{\sqrt{K_2}} = \large\frac{\sqrt{K_1}}{\sqrt{2K_1}}\: \: \: \: (K_2=2K_1)$
$ \Rightarrow \large\frac{\lambda_2}{ \lambda_1} = \large\frac{1}{\sqrt 2}$

 

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

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