The kinetic energy of an electron is increase | vedclass

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Question

(B) The kinetic energy $E$ is related to momentum $p$ by the formula $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.According to the de-Broglie h

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(B) The kinetic energy $E$ is related to momentum $p$ by the formula $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$.If the kinetic energy is increased by $2$ times,the new kinetic energy $E'$ becomes $E + 2E = 3E$.The new wavelength $\lambda'$ is $\lambda' = \frac{h}{\sqrt{2m(3E)}} = \frac{1}{\sqrt{3}} \lambda$.Therefore,the wavelength changes by a factor of $\frac{1}{\sqrt{3}}$.

The kinetic energy of an electron is increased by $2$ times,then the de-Broglie wavelength associated with it changes by a factor.

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