It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fracti

Q: Why do we assume the coefficient of restitution $e=1$?

The problem explicitly states that the collision is 'elastic'. In physics, a perfectly elastic collision is defined as one where kinetic energy is conserved, which mathematically translates to a coefficient of restitution $e = 1$.

Q: How do we know the masses of Deuterium and Carbon relative to the neutron?

A neutron has a mass of approximately $1 \text{ amu}$. Deuterium is an isotope of hydrogen with 1 proton and 1 neutron, giving it a mass of roughly $2 \text{ amu}$ ($M = 2m$). Carbon-12 has 6 protons and 6 neutrons, giving it a mass of roughly $12 \text{ amu}$ ($M = 12m$).

Q: Can I solve this without deriving the general formula $\frac{4mM}{(m+M)^2}$?

Yes, absolutely! You can plug in $M=2m$ directly into the momentum and restitution equations, solve for $v_1$, and then calculate the energy loss. You would then repeat the whole process for $M=12m$. However, deriving the general formula once saves a lot of calculation time.

Q: Why does the neutron lose more energy to Deuterium than to Carbon?

In elastic collisions, maximum kinetic energy is transferred when the masses of the two colliding bodies are exactly equal. Since the mass of Deuterium ($2m$) is much closer to the neutron's mass ($m$) than Carbon's mass ($12m$) is, the energy transfer is significantly higher.

Q: What would happen if the target mass was infinitely large?

If $M$ is infinitely large compared to $m$, the formula $\frac{4mM}{(m+M)^2}$ approaches $0$. Physically, the neutron would bounce back with almost the exact same speed it came in with ($v_1 \approx -v_0$), transferring practically zero kinetic energy to the massive wall.

Question

It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $P_d$; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $P_c$. The values of $P_d$ and $P_c$ are respectively

Options:

(A) (.89, .28)

(B) (.28, .89)

(C) (0, 0)

(D) (0, 1)

Sigma Percentile · Accepted Answer

### The Setup: A Neutron Meets a Nucleus

Imagine a microscopic game of billiards. A neutron, a tiny subatomic particle with mass $m$, is hurtling through space with an initial velocity $v_0$. It is on a direct collision course with a target nucleus of mass $M$, which is sitting perfectly still. This is a classic head-on, collinear collision.

When the neutron strikes the target, two fundamental laws of physics govern the outcome. First, the **Law of Conservation of Linear Momentum** dictates that the total momentum before the crash must equal the total momentum after. If the neutron bounces off with velocity $v_1$ and the target gets knocked forward with velocity $v_2$, we can write:

$mv_0 = mv_1 + Mv_2$

### The Master Equation: Fractional Energy Loss

The problem explicitly states that the collision is **elastic**. This is a crucial piece of information! It means that no kinetic energy is lost to heat or deformation. In mathematical terms, the coefficient of restitution $e$ is exactly $1$. This gives us a beautiful relationship between the velocities:

$e = \frac{v_2 - v_1}{v_0 - 0} = 1 \implies v_0 = v_2 - v_1$

Our ultimate goal is to find out how much kinetic energy the neutron loses. To do this, we need its final velocity $v_1$. By substituting $v_2 = v_0 + v_1$ into our momentum equation, we can solve for $v_1$:

$v_1 = \left( \frac{m - M}{m + M} \right) v_0$

Now, let's calculate the **fractional loss of kinetic energy**. This is simply the energy lost divided by the initial energy:

$\frac{\Delta K}{K_i} = \frac{\frac{1}{2}mv_0^2 - \frac{1}{2}mv_1^2}{\frac{1}{2}mv_0^2} = 1 - \left(\frac{v_1}{v_0}\right)^2$

Substituting our expression for $v_1$, we arrive at a highly powerful standard formula that every physics student should memorize:

$\frac{\Delta K}{K_i} = \frac{4mM}{(m+M)^2}$

### Case 1: The Deuterium Encounter

Let's apply our master formula to the first scenario. The neutron collides with a Deuterium nucleus. Deuterium is an isotope of hydrogen containing one proton and one neutron, making its mass approximately twice that of a single neutron ($M = 2m$).

Plugging this into our formula, we get the fractional energy loss $P_d$:

$P_d = \frac{4m(2m)}{(m+2m)^2} = \frac{8m^2}{9m^2} = \frac{8}{9}$

Calculating the decimal value, we find $P_d \approx 0.89$. The neutron loses a massive $89\%$ of its energy in a single bounce!

### Case 2: The Carbon Encounter

Now for the second scenario. The target is a Carbon nucleus. A standard Carbon-12 nucleus contains 6 protons and 6 neutrons, so its mass is roughly twelve times that of our incoming neutron ($M = 12m$).

Let's see how much energy is transferred this time. We calculate $P_c$:

$P_c = \frac{4m(12m)}{(m+12m)^2} = \frac{48m^2}{169m^2} = \frac{48}{169}$

This evaluates to $P_c \approx 0.28$. The neutron only loses about $28\%$ of its energy.

### The Grand Conclusion

This problem beautifully illustrates a core principle of nuclear physics and reactor design: **energy transfer is most efficient when the colliding masses are similar**. Because Deuterium's mass is much closer to the neutron's mass than Carbon's is, it acts as a far superior "moderator," stripping away the neutron's kinetic energy much more rapidly.

Our final values are $P_d = 0.89$ and $P_c = 0.28$, leading us straight to the correct option.

Select Answer:

Visualized Solution

$Elastic Collinear Collision$

$Momentum Conservation$

$Elastic Collision (e = 1)$

$Velocity of Neutron after Collision$

$Fractional Energy Loss$

$Collision with Deuterium$

$Collision with Carbon$

$Conclusion$

The Sigma Insight: Head-on Collision

The Setup

The Master Equation

Case 1

Case 2

The Grand Conclusion

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elastic Collinear Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Momentum Conservation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elastic Collision (e=1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Neutron after Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fractional Energy Loss

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Collision with Deuterium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Collision with Carbon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion