An alpha particle (4He) undergoes an elastic collision with a stationary protactinium nucleus (231Pa). What percent of the kinetic energy of the alpha particle is transferred to the protactinium nucleus? Assume the collision is one-dimensional.
To find the percent of kinetic energy transferred, we need to compare the initial kinetic energy of the alpha particle to the final kinetic energy of the protactinium nucleus.
1. Find the initial kinetic energy (KE_initial) of the alpha particle.
- KE_initial = (1/2) * m_alpha * v_alpha^2
- Where m_alpha is the mass of the alpha particle (4He) and is approximately 6.64 x 10^-27 kg.
- And v_alpha is the initial velocity of the alpha particle.
2. Find the final kinetic energy (KE_final) of the protactinium nucleus.
- Since the collision is elastic, both mass and kinetic energy are conserved.
- Notice that the protactinium nucleus is initially stationary, so its initial kinetic energy is zero.
- After the collision, the alpha particle will move with a different velocity (v_alpha_final), while the protactinium nucleus will start moving in the opposite direction with velocity (v_protactinium_final).
- So we have KE_final = (1/2) * m_alpha * v_alpha_final^2 + (1/2) * m_protactinium * v_protactinium_final^2
- Where m_protactinium is the mass of the protactinium nucleus (231Pa) and is approximately 3.88 x 10^-25 kg.
3. Calculate the percent of kinetic energy transferred.
- Percent of kinetic energy transferred = (KE_final / KE_initial) * 100
Note: Since this problem requires the knowledge of specific masses and velocities, please provide the initial velocity of the alpha particle to proceed with the calculation.