A radium nucleus (mass 226 u) at rest decays into a radon nucleus (symbol Rn, mass 222 u) and an alpha particle (symbol \(\alpha,\) mass 4 u). (a) Find the ratio of the speeds \(v_{\alpha} / v_{\mathrm{Rn}}\) after the decay. (b) Find the ratio of the magnitudes of the momenta \(p_{\alpha} / p_{\mathrm{Rn}^{*}}\) (c) Find the ratio of the kinetic energies \(K_{\alpha} / K_{\mathrm{Rn}} .\) (Note: "u" is a unit of mass; it is not necessary to convert it to kg.)

Considering that the radium nucleus is initially at rest and there are no external forces acting on the system, the total momentum before and after the decay should be equal: \(m_\mathrm{Ra}v_\mathrm{Ra} = m_\mathrm{Rn}v_\mathrm{Rn} + m_\alpha v_\alpha\) Initially, since the radium nucleus is at rest, \(m_\mathrm{Ra}v_\mathrm{Ra} = 0\). Thus, the equation becomes: \(0 = m_\mathrm{Rn}v_\mathrm{Rn} + m_\alpha v_\alpha\) We can rearrange the equation to find the ratio \(v_\alpha / v_\mathrm{Rn}\): \(v_\alpha / v_\mathrm{Rn} = -m_\mathrm{Rn} / m_\alpha\)

Using the masses given in the problem, we can substitute the values into the equation for the speed ratio: \(v_\alpha / v_\mathrm{Rn} = - (222\,\text{u}) / (4\,\text{u})\) \(v_\alpha / v_\mathrm{Rn} = -55.5\) The negative sign indicates that the velocities have opposite directions, as expected due to conservation of momentum. Now we have found the ratio of the speeds after decay: \(v_\alpha / v_\mathrm{Rn} = 55.5\)

Now let's calculate the ratio of the magnitudes of the momenta \(p_\alpha / p_\mathrm{Rn}\): Since \(p_\alpha = m_\alpha v_\alpha\) and \(p_\mathrm{Rn} = m_\mathrm{Rn} v_\mathrm{Rn}\), we can write the ratio as: \(p_\alpha / p_\mathrm{Rn} = (m_\alpha v_\alpha) / (m_\mathrm{Rn} v_\mathrm{Rn})\) By using the speed ratio we found earlier, we can simplify this expression: \(p_\alpha / p_\mathrm{Rn} = (m_\alpha / m_\mathrm{Rn}) (v_\alpha / v_\mathrm{Rn})\) \(p_\alpha / p_\mathrm{Rn} = (4\,\text{u} / 222\,\text{u}) (55.5)\) \(p_\alpha / p_\mathrm{Rn} = 1\) The ratio of the magnitudes of the momenta of the alpha particle and the radon nucleus is equal to 1.

Now let's calculate the ratio of the kinetic energies \(K_\alpha / K_\mathrm{Rn}\): Since \(K_\alpha = (1/2)m_\alpha v_\alpha^2\) and \(K_\mathrm{Rn} = (1/2)m_\mathrm{Rn} v_\mathrm{Rn}^2\), we can write the ratio as: \(K_\alpha / K_\mathrm{Rn} = [(1/2)m_\alpha v_\alpha^2] / [(1/2)m_\mathrm{Rn} v_\mathrm{Rn}^2]\) Using the speed ratio we found earlier, we can simplify this expression: \(K_\alpha / K_\mathrm{Rn} = (m_\alpha v_\alpha^2) / (m_\mathrm{Rn} v_\mathrm{Rn}^2)\) \(K_\alpha / K_\mathrm{Rn} = (m_\alpha / m_\mathrm{Rn}) (v_\alpha^2 / v_\mathrm{Rn}^2)\) \(K_\alpha / K_\mathrm{Rn} = (4\,\text{u} / 222\,\text{u}) (55.5^2)\) \(K_\alpha / K_\mathrm{Rn} = 13.75\) The ratio of the kinetic energies of the alpha particle and the radon nucleus is equal to 13.75. In conclusion: (a) The ratio of the speeds after decay is: \(v_\alpha / v_\mathrm{Rn} = 55.5\) (b) The ratio of the magnitudes of the momenta is: \(p_\alpha / p_\mathrm{Rn} = 1\) (c) The ratio of the kinetic energies is: \(K_\alpha / K_\mathrm{Rn} = 13.75\)