If {lambda _p} and {lambda _alpha } are the d | vedclass

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energ

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${\lambda _p} = 2{\lambda _\alpha }$

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.For a proton,$\lambda_p = \frac{h}{\sqrt{2m_p K}}$.For an $\alpha$-particle,the mass is $m_{\alpha} = 4m_p$. Given the kinetic energies are equal $(K_p = K_{\alpha} = K)$,we have $\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)K}} = \frac{h}{2\sqrt{2m_p K}}$.Comparing the two,we get $\lambda_{\alpha} = \frac{\lambda_p}{2}$,which implies $\lambda_p = 2\lambda_{\alpha}$.

If ${\lambda _p}$ and ${\lambda _\alpha }$ are the de-Broglie wavelengths associated with protons and $\alpha$-particles of equal kinetic energies,then:

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