If the kinetic energy of a free electron doub | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the

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$\frac{1}{\sqrt{2}}$

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(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.Since kinetic energy $K = \frac{p^2}{2m}$,we can express momentum as $p = \sqrt{2mK}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.This shows that $\lambda \propto \frac{1}{\sqrt{K}}$.If the kinetic energy $K$ is doubled to $K' = 2K$,the new wavelength $\lambda'$ becomes $\lambda' = \frac{h}{\sqrt{2m(2K)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mK}} = \frac{\lambda}{\sqrt{2}}$.Therefore,the wavelength changes by a factor of $\frac{1}{\sqrt{2}}$.

If the kinetic energy of a free electron doubles,its de-Broglie wavelength $\lambda$ changes by a factor of:

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