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(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the

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$2:1$

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

If a proton and an $\alpha$-particle have the same kinetic energy,what is the ratio of their de-Broglie wavelengths?

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