In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf/Ki , for a head-on elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is , where the atomic mass unit, m=1.009u , is defined as follows: 1 u = 1.66 x 10^(-27) kg.
A)An electron- M= 5.49 x 10^(-4)u .
B)A proton- M= 1.007u .
C)The nucleus of a lead atom M= 207.2u .
Use the formula for head-on elastic collisions that you will find at:
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
For the colliding neutron of mass m1,
Vf = Vi * (m1 - m2)/(m1 + m2)
m2 is the mass of the particle it collides with.
The largest mass loss will be in collisions with protons.
Well, when it comes to slowing down neutrons, it's like trying to slow down a toddler on a sugar rush - not an easy task! Now, let's calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy for each of the targets you mentioned:
A) An electron. Now, electrons might be small, but they can pack a punch (or a tiny kick). Given the mass of a neutron and the mass of an electron, the ratio of final kinetic energy to initial kinetic energy, Kf/Ki, will be less than 1. So, let's just say that an electron is like trying to slow down a neutron with a feather.
B) A proton. Ah, now we're talking about the big leagues! Protons have a similar mass to neutrons, so when they collide head-on, they can transfer a significant amount of their energy. Therefore, the ratio of final kinetic energy to initial kinetic energy, Kf/Ki, will be close to 1. Protons are like those enthusiastic cheerleaders, ready to boost the neutron's energy and keep the reaction going!
C) The nucleus of a lead atom. Whoa, talk about heavyweight! The nucleus of a lead atom is massive compared to a neutron. When a neutron smacks into it head-on, it's like hitting a brick wall! In this case, the ratio of final kinetic energy to initial kinetic energy, Kf/Ki, will be very small, approaching 0. Those lead nuclei are like the granddaddies of slowing down neutrons, taking away their energy faster than a whirlwind!
So, in summary, the electron is like a feather, the proton is like a cheerleader, and the lead nucleus is like a brick wall when it comes to slowing down neutrons.
To calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, we need to use the principle of conservation of momentum and kinetic energy in the collision. Let's go through each scenario:
A) Neutron colliding with an electron:
In an elastic collision, momentum and kinetic energy are conserved. Since the electron is much lighter than the neutron, we can neglect its initial momentum. Therefore, the neutron's final momentum will be equal to its initial momentum.
Initial momentum (pi) = Final momentum (pf)
From the conservation of momentum:
m_neutron * v_initial = m_neutron * v_final + m_electron * v_electron
Since the neutron is initially moving and the electron is stationary, we can set v_initial equal to the initial velocity of the neutron, and v_electron equal to zero.
m_neutron * v_initial = m_neutron * v_final
From the conservation of kinetic energy:
(1/2) * m_neutron * v_initial^2 = (1/2) * m_neutron * v_final^2 + (1/2) * m_electron * 0^2
Simplifying:
v_initial^2 = v_final^2
Taking the square root:
v_initial = v_final
This means the final kinetic energy of the neutron is equal to its initial kinetic energy.
Kf/Ki = 1
B) Neutron colliding with a proton:
Using the same approach as before, but considering the mass of the proton, we have:
m_neutron * v_initial = m_neutron * v_final + m_proton * v_proton
Since the proton is initially stationary, we can set v_initial equal to the initial velocity of the neutron, and v_proton equal to zero.
m_neutron * v_initial = m_neutron * v_final
v_initial = v_final
Kf/Ki = 1
C) Neutron colliding with a lead nucleus:
Using the same approach as before:
m_neutron * v_initial = m_neutron * v_final + m_lead * v_lead
Since the lead nucleus is initially stationary, we can set v_initial equal to the initial velocity of the neutron, and v_lead equal to zero.
m_neutron * v_initial = m_neutron * v_final
v_initial = v_final
Kf/Ki = 1
In all three scenarios, the ratio Kf/Ki is equal to 1. Therefore, in these head-on elastic collisions, the neutron's final kinetic energy is the same as its initial kinetic energy, regardless of the target particle.