The magnitude of the de-Broglie wavelength (l | vedclass

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(C) The de-Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we h

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$\lambda_\alpha, \lambda_{ n }, \lambda_{ p }, \lambda_{ e }$

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(C) The de-Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mE}}$.Given that the energy $E$ is the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Comparing the masses: $m_e Since $\lambda$ is inversely proportional to the square root of mass,the order of wavelengths will be $\lambda_\alpha Thus,the increasing order is $\lambda_\alpha, \lambda_n, \lambda_p, \lambda_e$.

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

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