A particle of charge q,mass m,and energy E ha | vedclass

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(D) The de-Broglie wavelength $\lambda$ of a particle with mass $m$ and kinetic energy $E$ is given by the formula:$\lambda = \frac{h}{\sqrt{2mE}}$For

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$\frac{\lambda}{2}$

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(D) The de-Broglie wavelength $\lambda$ of a particle with mass $m$ and kinetic energy $E$ is given by the formula:$\lambda = \frac{h}{\sqrt{2mE}}$For the second particle,the mass is $m' = 2m$ and the energy is $E' = 2E$. The charge $q$ does not affect the de-Broglie wavelength formula for a particle with a given kinetic energy.Substituting the new values into the formula:$\lambda' = \frac{h}{\sqrt{2m'E'}}$$\lambda' = \frac{h}{\sqrt{2(2m)(2E)}}$$\lambda' = \frac{h}{\sqrt{8mE}}$$\lambda' = \frac{h}{2\sqrt{2mE}}$Since $\lambda = \frac{h}{\sqrt{2mE}}$,we get:$\lambda' = \frac{\lambda}{2}$

$A$ particle of charge $q$,mass $m$,and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2q$,mass $2m$,and energy $2E$,the de-Broglie wavelength is:

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