If \(\propto\) -particle and proton are accelerated through the same potential difference, then the ratio of de Brogile wavelength of \(\propto\) -particle and proton is \(\ldots \ldots \ldots\) (A) \((1 / \sqrt{2})\) (B) \(\sqrt{2}\) (C) \(\\{1 /(2 \sqrt{2})\\}\) (D) \(2 \sqrt{2}\)

To calculate the momentum, we will use the relationship between kinetic energy and momentum: \(K.E. = \dfrac{p^2}{2m}\), where \(p\) is the momentum, and \(m\) is the mass of the particle. For the proton, \(p_p = \sqrt{2 \cdot m_p \cdot K.E.}\) For the alpha particle, \(p_\alpha = \sqrt{2 \cdot m_\alpha \cdot K.E.}\) #Step 3: Calculate the de Broglie wavelength for both particles#

To find the ratio as asked in the exercise, we need to divide the de Broglie wavelength of the alpha particle by the de Broglie wavelength of the proton: \(\dfrac{λ_\alpha}{λ_p} = \dfrac{(\dfrac{h}{p_\alpha})}{(\dfrac{h}{p_p})} = \dfrac{p_p}{p_\alpha} = \dfrac{\sqrt{2 \cdot m_p \cdot K.E.}}{\sqrt{2 \cdot m_\alpha \cdot K.E.}}\) Since \(m_\alpha = 4m_p\), we can substitute this into the equation: \(\dfrac{λ_\alpha}{λ_p} = \dfrac{\sqrt{2 \cdot m_p \cdot K.E.}}{\sqrt{2 \cdot (4m_p) \cdot K.E.}} = \dfrac{\sqrt{2 \cdot m_p \cdot K.E.}}{\sqrt{8 \cdot m_p \cdot K.E.}} = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}\) The ratio of the de Broglie wavelengths of the alpha particle and the proton is \(\dfrac{1}{2}\), which corresponds to answer (C).