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A proton and an $$\alpha$$-particle are accelerated from rest by $$2$$ V and $$4$$ V potentials, respectively. The ratio of their de-Broglie wavelength is :
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$$
$$\lambda \propto \frac{1}{\sqrt{mqV}}$$
$$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_p q_p V_p}}$$
$$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{(4m_p)(2q_p)(4)}{(m_p)(q_p)(2)}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$
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