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

Vedclass · Accepted Answer

$\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$

Vedclass · Answer

(B) The De-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.Given that the kinetic energy $K$ is the same for all particles,$\lambda \propto \frac{1}{\sqrt{m}}$.We know the masses of the particles are related as $m_{\alpha} > m_{p} > m_{e}$.Therefore,the relationship between their wavelengths is $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$.

An $\alpha$-particle,a proton,and an electron have the same kinetic energy. Which one of the following is correct in the case of their De-Broglie wavelength?

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