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%.
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?
1 answer