Question

A 238 92U nucleus is moving in the x-direction at 5.0 × 105 m/s when it decays into an alpha particle 4 2He and a 234 90Th. If the alpha particle moves off at 25.4° above the x axis with a speed of 1.4 × 107 m/s, what is the recoil velocity of the thorium nucleus? Assume the uranium-thorium-alpha system is isolated; you may also assume the particles are pointlike.

Answer 1

Answer:

v_th = 3.1 * 10^5 m/s

Explanation:

Given:

- mass of 238-Uranium m_u = 3.952 *10^-25 kg

- mass of alpha particle m_a = 6.64 * 10^-27 kg

- mass of thorium particle m_th = 3.885*10^-25 kg

- velocity of 238-Uranium v_u = 5.0 *10^5 m/s

- velocity of alpha particle v_a = 1.4 *10^7 m/s

Find:

- The recoil velocity of the thorium particle.

Solution:

- To solve this problem we will use conservation of momentum in both x and y direction.

- Momentum conservation in x-direction:

                                        P_i = P_f

                  m_u*v_u = m_a*v_a*cos(Q) + m_th*v_th,x

where v_th,x is the x component of thorium velocity:

                 P_i = 3.952 *10^-25 * 5.0 *10^5 = 1.976*10^-19

      P_f = 6.64 * 10^-27*1.4 *10^7*cos(25.4) + 3.885*10^-25*v_th,x

                 P_f = 8.3974*10^-20 + 3.885*10^-25*v_th,x

Hence,

             1.976*10^-19 = 9.296*10^-20 + 3.885*10^-25*v_th,x

                         v_th,x = 2.92473 * 10^5 m/s        

- Momentum conservation in y-direction:

                                        P_i = P_f

                         0 = m_a*v_a*sin(Q) + m_th*v_th,y

where v_th,x is the x component of thorium velocity:

                         v_th,y = m_a*v_a*sin(Q) / m_th    

        v_th,y = 6.64 * 10^-27*1.4 *10^7*sin(25.4) / 3.885*10^-25

Hence,

                            v_th,y = 1.02635 * 10^5 m/s        

- The magnitude of recoil velocity:

                        v_th = sqrt ( v_th,x ^2 + v_th,y ^2 )

            v_th = sqrt ( (2.92473 * 10^5)^2 + (1.02635 * 10^5)^2 )

                                  v_th = 3.1 * 10^5 m/s