A neutron in a reactor makes an elastic head-on collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is about 12.5 the mass of the neutron.
What fraction of the neutron's kinetic energy is transferred to the atomic nucleus?
To solve this problem, we need to apply the conservation of kinetic energy and momentum.
Let's denote the mass of the neutron as m_neutron and the mass of the atomic nucleus as m_nucleus. According to the problem, m_nucleus = 12.5 * m_neutron.
Before the collision, only the neutron is moving, and the nucleus is at rest. Therefore, the initial momentum of the system is given by:
P_initial = m_neutron * v_neutron, where v_neutron is the initial velocity of the neutron.
After the collision, the neutron and the nucleus move away from each other with velocities v_neutron' and v_nucleus', respectively. The final momentum is given by:
P_final = m_neutron * v_neutron' + m_nucleus * v_nucleus'.
According to the conservation of momentum, the total initial momentum and the total final momentum should be equal:
P_initial = P_final
m_neutron * v_neutron = m_neutron * v_neutron' + m_nucleus * v_nucleus'
Now let's consider the conservation of kinetic energy. The kinetic energy of the system before the collision is given by:
KE_initial = 0.5 * m_neutron * v_neutron^2
Similarly, the kinetic energy of the system after the collision is given by:
KE_final = 0.5 * m_neutron * v_neutron'^2 + 0.5 * m_nucleus * v_nucleus'^2
According to the conservation of kinetic energy, the total initial kinetic energy and the total final kinetic energy should be equal:
KE_initial = KE_final
0.5 * m_neutron * v_neutron^2 = 0.5 * m_neutron * v_neutron'^2 + 0.5 * m_nucleus * v_nucleus'^2
Now we have a system of equations to solve for v_neutron' and v_nucleus'. Let's solve it.
First, we divide the second equation by the first equation to eliminate the constant terms and mass of the neutron:
(v_neutron')^2 = (v_neutron)^2 + (m_nucleus / m_neutron) * (v_nucleus')^2
Plugging in the value of m_nucleus / m_neutron = 12.5, we get:
(v_neutron')^2 = (v_neutron)^2 + 12.5 * (v_nucleus')^2
Now we can substitute this equation into the conservation of momentum equation and solve for v_nucleus':
m_neutron * v_neutron = m_neutron * v_neutron' + m_nucleus * v_nucleus'
m_neutron * v_neutron = m_neutron * v_neutron' + (12.5 * m_neutron) * v_nucleus'
v_nucleus' = (v_neutron - v_neutron') / 12.5
Finally, let's plug in the expression for v_nucleus' into the equation for kinetic energy conservation:
0.5 * m_neutron * v_neutron^2 = 0.5 * m_neutron * v_neutron'^2 + 0.5 * (12.5 * m_neutron) * [(v_neutron - v_neutron') / 12.5]^2
Simplifying the equation, we get:
v_neutron'^2 = 0.04 * (v_neutron - v_neutron')^2
Canceling out the common terms, we are left with:
1 = 0.04 * (v_neutron - v_neutron')^2
Taking the square root of both sides, we get:
1 = 0.2 * (v_neutron - v_neutron')
Now we can solve for the ratio of v_neutron' to v_neutron:
v_neutron' / v_neutron = 0.8
To calculate the fraction of the neutron's kinetic energy transferred to the atomic nucleus, we can use the formula:
Fraction = (KE_initial - KE_final) / KE_initial
Plugging in the values, we have:
Fraction = (0.5 * m_neutron * v_neutron^2 - 0.5 * m_neutron * v_neutron'^2) / (0.5 * m_neutron * v_neutron^2)
= (v_neutron^2 - v_neutron'^2) / v_neutron^2
= (v_neutron^2 - (v_neutron^2 * 0.8^2)) / v_neutron^2
= (1 - 0.64) / 1
= 0.36
Therefore, the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 0.36 or 36%.