Question 17

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $$\lambda$$. An alpha particle having certain kinetic energy has the same de-Broglie wavelength $$\lambda$$. The ratio of kinetic energy of proton and that of alpha particle is :

Using the de Broglie relation: $$\lambda = \frac{h}{\sqrt{2mK}} \implies K = \frac{h^2}{2m\lambda^2}$$

Given: $$\lambda_p = \lambda_\alpha = \lambda$$

$$\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p}$$

$$m_\alpha = 4m_p$$

$$\frac{K_p}{K_\alpha} = \frac{4m_p}{m_p} = \frac{4}{1}$$

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