The de-Broglie wavelength of a proton and \(\alpha\) - particle is same. The ratio of their velocities will be.......... ( \(\alpha\) particle is the He- nucleus, having two protons and two neutrons. Thus, its mass \(\mathrm{M}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\) where \(\mathrm{m}_{\mathrm{p}}\) is the mass of the proton.) (A) \(1: 4\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)

For a particle with mass m and velocity v, the de-Broglie wavelength is given by the formula, \(\lambda = \frac{h}{mv}\), where h is Planck's constant. For a proton, with mass \(m_p\) and velocity \(v_p\), its de-Broglie wavelength \(\lambda_p = \frac{h}{m_p v_p}\). For an \(\alpha\)-particle, with mass \(M_\alpha = 4m_p\) and velocity \(v_\alpha\), its de-Broglie wavelength \(\lambda_\alpha = \frac{h}{(4m_p) v_\alpha}\).

We are given that the de-Broglie wavelengths of the proton and the \(\alpha\)-particle are the same, i.e., \(\lambda_p = \lambda_\alpha\). Substitute the formulas from Step 1: \(\frac{h}{m_pv_p} = \frac{h}{(4m_p)v_\alpha}\) Now, we need to find the ratio of the velocities, \(\frac{v_p}{v_\alpha}\). Divide both sides by \(h\) and rearrange the equation: \(\frac{v_p}{v_\alpha} = \frac{4m_p v_\alpha}{m_pv_p}\) To solve for the velocity ratio, multiply both sides by \(v_p\) and divide both sides by \(4m_pv_\alpha\): \(\frac{v_p}{v_\alpha} \times \frac{v_p}{4m_p} \times \frac{1}{v_\alpha} = \frac{4m_p v_\alpha}{m_p v_p} \times \frac{v_p}{4m_p} \times \frac{1}{v_\alpha}\) This simplifies to: \(\frac{v_p^2}{4v_\alpha^2} = 1\) Take the square-root of both sides: \(\frac{v_p}{2v_\alpha} = 1\)

We have found that \(\frac{v_p}{2v_\alpha} = 1\), which implies: \(v_p = 2v_\alpha\) Now, divide both sides by \(v_\alpha\): \(\frac{v_p}{v_\alpha} = 2\) So, the ratio of their velocities is \(2:1\), which corresponds to choice (C).