A heavy nucleus N, at rest, undergoes fission N - P + Q, where P and Q are two lighter nuclei. Let

Q: Why is the initial momentum zero?

The problem explicitly states that the heavy nucleus N is 'at rest' before the fission occurs. Therefore, its initial velocity is zero, making the initial momentum zero.

Q: Why does mass defect release energy?

According to Einstein's mass-energy equivalence principle ($E=mc^2$), mass and energy are interchangeable. The missing mass ($\delta$) is converted into the kinetic energy of the fragments.

Q: Why is kinetic energy inversely proportional to mass here?

Because the fragments have equal and opposite momenta ($p$), we can use the formula $K = p^2/2m$. Since $p$ is constant for both, $K \propto 1/m$.

Q: What is reduced mass $\mu$?

Reduced mass is an effective inertial mass appearing in the two-body problem. It simplifies the math when dealing with two interacting bodies, defined as $\mu = \frac{m_1 m_2}{m_1 + m_2}$.

Q: Why is option B incorrect?

Because kinetic energy is inversely proportional to mass, the lighter fragment gets a larger share of the energy. The energy of P is proportional to the mass of Q, so $E_P = \frac{M_Q}{M_P + M_Q} \delta c^2$.

Question

A heavy nucleus N, at rest, undergoes fission $	ext{N} ightarrow 	ext{P} + 	ext{Q}$, where P and Q are two lighter nuclei. Let $\delta = M_N - M_P - M_Q$, where $M_P$, $M_Q$ and $M_N$ are the masses of P, Q and N, respectively. $E_P$ and $E_Q$ are the kinetic energies of P and Q, respectively. The speed of P and Q are $v_P$ and $v_Q$, respectively. If $c$ is the speed of light, which of the following statement(s) is(are) correct?

Options:

(A) $E_P + E_Q = c^2 \delta$

(B) $E_P = \left(\frac{M_P}{M_P + M_Q}ight) c^2 \delta$

(C) $\frac{v_P}{v_Q} = \frac{M_Q}{M_P}$

(D) The magnitude of momentum for P as well as Q is $c\sqrt{2\mu\delta}$, where $\mu = \frac{M_P M_Q}{M_P + M_Q}$

Sigma Percentile · Accepted Answer

### The Anatomy of Fission

Imagine a heavy, unstable nucleus, let's call it $\mathbf{N}$, sitting perfectly at rest. Suddenly, the strong nuclear forces can no longer hold it together, and it violently splits into two lighter fragments, $\mathbf{P}$ and $\mathbf{Q}$. Because the parent nucleus was initially at rest, the two fragments must fly apart in exactly opposite directions to ensure that the total momentum of the universe remains undisturbed.

But where does the immense kinetic energy of these flying fragments come from?

### The Energy Source: Mass Defect

This is where Einstein's legendary mass-energy equivalence principle, $E = mc^2$, takes the center stage. If you were to carefully weigh the initial nucleus $\mathbf{N}$ and then weigh the two resulting fragments $\mathbf{P}$ and $\mathbf{Q}$, you would notice something astonishing: some mass has vanished!

This missing mass is known as the **mass defect**, denoted by $\delta$:

$$\delta = M_N - M_P - M_Q$$

This "lost" mass hasn't truly disappeared; it has been converted into pure kinetic energy. The total energy released, often called the Q-value of the reaction, is given by:

$$Q = \delta c^2$$

Since the initial nucleus was at rest, this entire energy is distributed as the kinetic energy of the two fragments. Therefore, the sum of their kinetic energies must equal the total energy released:

$$E_P + E_Q = \delta c^2$$

This elegant application of the conservation of energy immediately confirms that **Option (A)** is correct.

### The Dance of Momentum

Now, let's look at the system through the lens of momentum. Before the fission, the heavy nucleus $\mathbf{N}$ was at rest, meaning the initial momentum was exactly zero. According to the law of conservation of linear momentum, the final total momentum must also be zero.

$$\vec{p}_{initial} = 0 \implies \vec{p}_P + \vec{p}_Q = 0$$

This implies that the two fragments must have momenta that are equal in magnitude but opposite in direction. Let's denote this common magnitude of momentum as $p$:

$$|\vec{p}_P| = |\vec{p}_Q| = p$$

Since momentum is the product of mass and velocity ($p = mv$), we can write:

$$M_P v_P = M_Q v_Q$$

Rearranging this equation gives us the ratio of their speeds:

$$\frac{v_P}{v_Q} = \frac{M_Q}{M_P}$$

This tells us that the lighter fragment will fly away much faster than the heavier one, perfectly confirming that **Option (C)** is correct.

### Distributing the Energy Spoils

To find out exactly how much kinetic energy each fragment gets, it is incredibly useful to express kinetic energy in terms of momentum. The standard formula $K = \frac{1}{2}mv^2$ can be rewritten as:

$$K = \frac{p^2}{2m}$$

Because both fragments share the exact same magnitude of momentum $p$, their kinetic energy is **inversely proportional** to their mass ($K \propto \frac{1}{m}$). This means the lighter fragment gets the lion's share of the energy!

To find the exact kinetic energy of fragment $\mathbf{P}$, we divide the total energy in the inverse ratio of their masses:

$$E_P = \left(\frac{M_Q}{M_P + M_Q}\right) \delta c^2$$

Notice that the numerator contains $M_Q$, not $M_P$. This reveals that **Option (B)** is incorrect.

### The Final Calculation: Finding Momentum

Finally, let's determine the exact mathematical expression for the momentum $p$. We can substitute our momentum-based kinetic energy formulas back into the total energy equation:

$$\frac{p^2}{2M_P} + \frac{p^2}{2M_Q} = \delta c^2$$

Factoring out $\frac{p^2}{2}$, we get:

$$\frac{p^2}{2} \left(\frac{1}{M_P} + \frac{1}{M_Q}\right) = \delta c^2$$

$$\frac{p^2}{2} \left(\frac{M_P + M_Q}{M_P M_Q}\right) = \delta c^2$$

The term inside the parenthesis is the reciprocal of the **reduced mass**, $\mu$, which is defined as $\mu = \frac{M_P M_Q}{M_P + M_Q}$. Substituting $\mu$ into our equation simplifies it beautifully:

$$\frac{p^2}{2\mu} = \delta c^2$$

Solving for $p$, we take the square root of both sides:

$$p = c\sqrt{2\mu\delta}$$

This flawless derivation confirms that the magnitude of momentum for both fragments is indeed $c\sqrt{2\mu\delta}$, making **Option (D)** correct.

Select Answer:

Visualized Solution

\text{Fission Process}

\text{Q-value of the Reaction}

\text{Conservation of Energy}

\text{Conservation of Momentum}

\text{Ratio of Speeds}

\text{Kinetic Energy and Momentum}

\text{Kinetic Energy of P}

\text{Calculating Momentum } p

\text{Final Momentum Expression}

The Sigma Insight: Nucleus and Nuclear Reaction

The Anatomy of Fission

The Energy Source

The Dance of Momentum

Distributing the Energy Spoils

The Final Calculation

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Select Answer:

Visualized Solution

\text{Fission Process}

\text{Q-value of the Reaction}

\text{Conservation of Energy}

\text{Conservation of Momentum}

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\text{Kinetic Energy and Momentum}

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The Anatomy of Fission

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The Dance of Momentum

Distributing the Energy Spoils

The Final Calculation

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