A proton and an alpha-particle are accelerate | vedclass

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of t

Vedclass · Accepted Answer

$2:1$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E_k$ is the kinetic energy.Since both particles have the same kinetic energy $E_k$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,i.e.,$m_{\alpha} = 4m_p$.Substituting this value,we get $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2$.Thus,the ratio $\lambda_p : \lambda_{\alpha}$ is $2:1$.

$A$ proton and an $\alpha$-particle are accelerated from rest to the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_{p} : \lambda_{\alpha}$ is:

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