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

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$1:2$

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(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.Given: $K_p : K_\alpha = 16 : 1$. The mass of an $\alpha$-particle $(m_\alpha)$ is approximately $4$ times the mass of a proton $(m_p)$,so $m_\alpha = 4m_p$.The ratio of wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{2m_\alpha K_\alpha}}{\sqrt{2m_p K_p}} = \sqrt{\frac{m_\alpha}{m_p} \cdot \frac{K_\alpha}{K_p}}$.Substituting the values: $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p}{m_p} \cdot \frac{1}{16}} = \sqrt{4 \cdot \frac{1}{16}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio is $1:2$.

The ratio of kinetic energies of a proton and an $\alpha$-particle is $16:1$. What is the ratio of their associated de Broglie wavelengths?

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