A particle has a de Broglie wavelength lambda | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the velocity $v$ becomes $2v$ and the kinetic energy $E = \frac{1}{2}mv^

Vedclass · Accepted Answer

$\frac{\lambda}{2}$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the velocity $v$ becomes $2v$ and the kinetic energy $E = \frac{1}{2}mv^2$ becomes $2E$.Let the new wavelength be $\lambda'$.Using the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,we have $\lambda \propto \frac{1}{\sqrt{E}}$.If the energy $E$ is doubled to $2E$,the new wavelength $\lambda' = \frac{\lambda}{\sqrt{2}}$.However,the question specifies that both velocity and energy are doubled. Since $E = \frac{1}{2}mv^2$,if $v$ is doubled,$E$ must become $4E$. If the problem states $E$ is doubled independently,it implies a change in mass or a specific constraint. Assuming the standard de Broglie relation $\lambda = \frac{h}{mv}$:If $v' = 2v$ and $m$ is constant,then $\lambda' = \frac{h}{m(2v)} = \frac{\lambda}{2}$.Given the options provided,the correct physical interpretation for a particle with constant mass is $\lambda' = \frac{\lambda}{2}$.

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

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