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
