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The de-Broglie wavelength of a free electron with kinetic energy 'E' is $ \lambda$. If the kinetic energy of the electron is doubled, the de-Broglie wavelength is

$\begin {array} {1 1} (1)\;\large\frac{\lambda}{\sqrt 2} & \quad (2)\;\sqrt 2 \lambda \\ (3)\;\large\frac{\lambda}{2} & \quad (4)\;2\lambda \end {array}$

1 Answer

(1) $ \large\frac{\lambda}{\sqrt 2}$
answered Nov 7, 2013 by pady_1
 

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