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(D) The de Broglie wavelength $\lambda$ is related to linear momentum $p$ by the equation $\lambda = \frac{h}{p}$.Given that the linear momentum of th

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The de Broglie wavelength of the proton and electron are the same.

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(D) The de Broglie wavelength $\lambda$ is related to linear momentum $p$ by the equation $\lambda = \frac{h}{p}$.Given that the linear momentum of the proton $(p_p)$ and the electron $(p_e)$ are equal,i.e.,$p_p = p_e$.Since $\lambda = \frac{h}{p}$,if $p$ is the same for both particles,then their de Broglie wavelengths must also be equal.Therefore,$\lambda_p = \lambda_e$.Regarding kinetic energy $K$,it is given by $K = \frac{p^2}{2m}$.Since $p$ is constant,$K \propto \frac{1}{m}$.Because the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,the kinetic energy of the proton will be less than that of the electron.Thus,option $D$ is correct.

If a proton and an electron have the same linear momentum,then compared to the electron:

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