The ratio of de Broglie wavelengths of a prot | vedclass

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Since the energy $E$ is the same for both particles,we have $

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$2 : 1$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Since the energy $E$ is the same for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,the ratio of the wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}}$.Given that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton $(m_\alpha = 4m_p)$,we get:$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = \frac{2}{1}$.Thus,the ratio is $2 : 1$.

The ratio of de Broglie wavelengths of a proton and an $\alpha$-particle having the same energy is ............

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