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(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particl

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$\beta$-particle

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(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.Given that the velocity $v$ is the same for all particles,the relationship becomes $\lambda \propto \frac{1}{m}$.This means that the particle with the smallest mass will have the maximum de-Broglie wavelength.Comparing the masses: $m_{\beta} Since the $\beta$-particle (an electron) has the smallest mass among the given options,it will have the maximum de-Broglie wavelength.

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

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