A fission reaction is given by 236_92 U - 140_54 Xe + 94_38 Sr + x + y, where x and y are two parti

Q: Why do we ignore the momentum of the neutrons when calculating the kinetic energies of the fragments?

Momentum is given by $P = \sqrt{2mK}$. The neutrons have a very small mass (1 amu) and a small kinetic energy (2 MeV) compared to the heavy fragments (masses 140 and 94 amu, energies around 100 MeV). Therefore, their momentum is negligible, and we can assume the heavy fragments perfectly balance each other's momentum.

Q: Why is the kinetic energy inversely proportional to the mass of the fragments?

Because the initial Uranium nucleus was at rest, the two heavy fragments must fly apart with equal and opposite momenta to conserve momentum (ignoring the tiny neutrons). Since $K = \frac{P^2}{2m}$, and $P$ is the same for both, the kinetic energy $K$ is inversely proportional to the mass $m$.

Q: How exactly do we calculate the Q-value of the reaction?

The Q-value is the total energy released. It is calculated by subtracting the total binding energy of the reactants from the total binding energy of the products. Total binding energy is the binding energy per nucleon multiplied by the total number of nucleons (mass number).

Q: Why does the lighter fragment get more kinetic energy?

To conserve momentum, the lighter fragment must travel at a much higher velocity than the heavier fragment. Since kinetic energy depends on the square of the velocity ($K = \frac{1}{2}mv^2$), the higher velocity of the lighter fragment results in it having a larger share of the total kinetic energy.

Q: What would happen if the Uranium nucleus was not initially at rest?

If the Uranium nucleus was moving, the total initial momentum would not be zero. We would have to use vector addition to conserve momentum, which would make the calculation significantly more complex and the fragments would not necessarily fly apart in exactly opposite directions.

Question

A fission reaction is given by $_{92}^{236}	ext{U} ightarrow \, _{54}^{140}	ext{Xe} + \, _{38}^{94}	ext{Sr} + x + y$,
where $x$ and $y$ are two particles. Considering $_{92}^{236}	ext{U}$ to be at rest, the kinetic energies of the products are denoted by $K_{	ext{Xe}}$, $K_{	ext{Sr}}$, $K_x$(2 MeV) and $K_y$(2 MeV), respectively. Let the binding energies per nucleon of $_{92}^{236}	ext{U}$, $_{54}^{140}	ext{Xe}$ and $_{38}^{94}	ext{Sr}$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are

Options:

(A) $x = n$, $y = n$, $K_{	ext{Sr}} = 129	ext{ MeV}$, $K_{	ext{Xe}} = 86	ext{ MeV}$

(B) $x = p$, $y = e^-$, $K_{	ext{Sr}} = 129	ext{ MeV}$, $K_{	ext{Xe}} = 86	ext{ MeV}$

(C) $x = p$, $y = n$, $K_{	ext{Sr}} = 129	ext{ MeV}$, $K_{	ext{Xe}} = 86	ext{ MeV}$

(D) $x = n$, $y = n$, $K_{	ext{Sr}} = 86	ext{ MeV}$, $K_{	ext{Xe}} = 129	ext{ MeV}$

Sigma Percentile · Accepted Answer

Nuclear fission is not just a reaction; it is a spectacular display of mass converting into pure, unadulterated energy. In this problem, we are witnessing the splitting of a Uranium-236 nucleus into Xenon-140 and Strontium-94. But wait, there are two mysterious particles, $x$ and $y$, that also emerge from this chaotic event. Let's embark on a journey to uncover their identities and trace the flow of energy.

### Identifying the Unknown Particles

In the quantum realm, the laws of conservation are absolute. The total number of nucleons (mass number) and the total charge (atomic number) must remain constant before and after the fission.

Let's balance the mass numbers ($A$):
$$236 = 140 + 94 + A_x + A_y$$
$$A_x + A_y = 2$$

Now, let's balance the atomic numbers ($Z$):
$$92 = 54 + 38 + Z_x + Z_y$$
$$Z_x + Z_y = 0$$

Since both particles have a combined mass number of 2 and a combined atomic number of 0, it is clear that $x$ and $y$ are both neutrons ($_{0}^{1}n$).

### The Q-Value: Unleashing the Energy

The energy released in a nuclear reaction, known as the Q-value, is the difference between the total binding energy of the products and the reactants.

$$Q = BE_{	ext{products}} - BE_{	ext{reactants}}$$
$$Q = (140 	imes 8.5 + 94 	imes 8.5) - (236 	imes 7.5)$$
$$Q = 1989 - 1770 = 219 	ext{ MeV}$$

This $219 	ext{ MeV}$ of energy doesn't just vanish; it manifests as the kinetic energy of the newly formed particles.

### Distributing the Kinetic Energy

We know the total kinetic energy is shared among the Xenon nucleus, the Strontium nucleus, and the two neutrons. The problem states that each neutron carries away $2 	ext{ MeV}$ of kinetic energy.

$$K_{	ext{Xe}} + K_{	ext{Sr}} + K_x + K_y = Q$$
$$K_{	ext{Xe}} + K_{	ext{Sr}} + 2 + 2 = 219$$
$$K_{	ext{Xe}} + K_{	ext{Sr}} = 215 	ext{ MeV}$$

Now, how is this $215 	ext{ MeV}$ divided between Xenon and Strontium? This is where the **Conservation of Linear Momentum** steps in. The initial Uranium nucleus was at rest, so the total initial momentum was zero.

Here is a crucial approximation: The momentum of the neutrons is extremely small compared to the massive fragments because both their mass and kinetic energy are tiny ($P = \sqrt{2mK}$). Therefore, we can safely assume that the Xenon and Strontium nuclei fly apart with approximately equal and opposite momenta.

$$|P_{	ext{Xe}}| \approx |P_{	ext{Sr}}|$$

Since kinetic energy $K = \frac{P^2}{2m}$, and the momenta are equal, the kinetic energy is inversely proportional to the mass. The lighter fragment will always zip away with a larger share of the energy!

$$\frac{K_{	ext{Sr}}}{K_{	ext{Xe}}} = \frac{m_{	ext{Xe}}}{m_{	ext{Sr}}} = \frac{140}{94}$$

Let's calculate the exact shares:
$$K_{	ext{Sr}} = \frac{140}{140+94} 	imes 215 \approx 129 	ext{ MeV}$$
$$K_{	ext{Xe}} = \frac{94}{140+94} 	imes 215 \approx 86 	ext{ MeV}$$

And there we have it! The lighter Strontium nucleus takes the lion's share of the energy ($129 	ext{ MeV}$), while the heavier Xenon nucleus takes the rest ($86 	ext{ MeV}$). This perfectly matches option (a).

Select Answer:

Visualized Solution

$Fission Reaction$

$Conservation of Nucleons$

$Q-Value Calculation$

$Kinetic Energy of Fragments$

$Momentum Conservation$

$Individual Kinetic Energies$

$Why Ignore Neutron Momentum?$

The Sigma Insight: Nucleus and Nuclear Reaction

Identifying the Unknown Particles

The Q-Value

Distributing the Kinetic Energy

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

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

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

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