If the kinetic energy of a free electron is d | vedclass

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.From this relation,we can see that $\la

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$1/\sqrt{2}$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$.If the kinetic energy $E$ is doubled (i.e.,$E' = 2E$),the new wavelength $\lambda'$ will be $\lambda' = \frac{h}{\sqrt{2m(2E)}} = \frac{1}{\sqrt{2}} 	imes \frac{h}{\sqrt{2mE}} = \frac{1}{\sqrt{2}} \lambda$.Therefore,the de Broglie wavelength becomes $1/\sqrt{2}$ times the original wavelength.

If the kinetic energy of a free electron is doubled,its de Broglie wavelength becomes ........... times.

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