If the kinetic energy of a free electron doub | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.From this expression,it is clear that $\lambda \propto \frac{1}{\sqrt{E}}$.If the kinetic energy $E$ becomes $E' = 2E$,the new wavelength $\lambda'$ is given by $\lambda' = \frac{h}{\sqrt{2m(2E)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mE}}$.Therefore,$\lambda' = \frac{1}{\sqrt{2}} \lambda$.Thus,the wavelength changes by a factor of $1/\sqrt{2}$.

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

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