The user wants me to solve a multiple-choice question from Physics, specifically from the topic of Atoms and Nuclei.

Question Analysis:

• A heavy nucleus N is at rest.
• It undergoes fission into two lighter nuclei P and Q.
• The reaction is: N $\to$ P + Q.
• Mass defect is given as $\delta = M_N - M_P - M_Q$.
• $M_N, M_P, M_Q$ are the masses of the nuclei.
• $E_P, E_Q$ are the kinetic energies of P and Q.
• $v_P, v_Q$ are the speeds of P and Q.
• c is the speed of light.

I need to check the correctness of four statements.

Step 1: Analyze the conservation of energy. The fission process releases energy. The source of this energy is the mass defect. The initial energy of the system is the rest mass energy of nucleus N, which is $M_N c^2$. Since N is at rest, its kinetic energy is zero. The final energy of the system is the sum of the rest mass energies and kinetic energies of the products P and Q. Final energy = $(M_P c^2 + E_P) + (M_Q c^2 + E_Q)$.

According to the principle of conservation of energy (or mass-energy equivalence): Initial Energy = Final Energy $M_N c^2 = M_P c^2 + M_Q c^2 + E_P + E_Q$

Rearranging the terms: $E_P + E_Q = M_N c^2 - M_P c^2 - M_Q c^2$ $E_P + E_Q = (M_N - M_P - M_Q) c^2$

We are given that $\delta = M_N - M_P - M_Q$. So, the total kinetic energy released, which is also called the Q-value of the reaction, is: $E_P + E_Q = \delta c^2$

This matches option A. So, statement A is correct.

Step 2: Analyze the conservation of momentum. The initial nucleus N is at rest. So, the initial momentum of the system is zero. $\vec{p}_{initial} = 0$

After the fission, the two nuclei P and Q move in opposite directions to conserve momentum. The final momentum of the system is the vector sum of the momenta of P and Q. $\vec{p}_{final} = \vec{p}_P + \vec{p}_Q$

According to the principle of conservation of linear momentum: $\vec{p}_{initial} = \vec{p}_{final}$ $0 = \vec{p}_P + \vec{p}_Q$ $\vec{p}_P = - \vec{p}_Q$

This means the momenta of P and Q are equal in magnitude and opposite in direction. Let the magnitude of their momenta be $p$. $|\vec{p}_P| = |\vec{p}_Q| = p$

The momentum of P is $p_P = M_P v_P$. The momentum of Q is $p_Q = M_Q v_Q$. (Here, we are using non-relativistic momentum. This is a standard approximation for fission fragments.)

So, $M_P v_P = M_Q v_Q$. From this, we can find the ratio of their speeds: ${{{v_P}} \over {{v_Q}}} = {{{M_Q}} \over {{M_P}}}$

This matches option C. So, statement C is correct.

Step 3: Analyze the individual kinetic energies. We know the total kinetic energy from step 1: $E_{total} = E_P + E_Q = \delta c^2$. And we know the relationship between their momenta from step 2: $p_P = p_Q = p$.

The kinetic energy of a particle with mass M and momentum p is given by $E = {p^2 \over 2M}$. So, $E_P = {p^2 \over 2M_P}$ and $E_Q = {p^2 \over 2M_Q}$.

From these relations, we can see that $E_P M_P = {p^2 \over 2}$ and $E_Q M_Q = {p^2 \over 2}$. Therefore, $E_P M_P = E_Q M_Q$. This gives the ratio of their kinetic energies: ${E_P \over E_Q} = {M_Q \over M_P}$ The lighter nucleus gets more kinetic energy.

Now let's check option B. It gives an expression for $E_P$. We have two equations:

1. $E_P + E_Q = \delta c^2$
2. $E_Q = E_P {M_P \over M_Q}$

Substitute (2) into (1): $E_P + E_P {M_P \over M_Q} = \delta c^2$ $E_P (1 + {M_P \over M_Q}) = \delta c^2$ $E_P ({{M_Q + M_P} \over M_Q}) = \delta c^2$ $E_P = \left({M_Q \over M_P + M_Q}\right) \delta c^2$

Now let's compare this with option B: Option B: ${E_P} = \left( {{{{M_P}} \over {{M_P} + {M_Q}}}} \right){c^2}\delta$ Our derived expression is different. Option B is incorrect.

Step 4: Analyze the magnitude of momentum. We need to find the magnitude of the momentum, $p$. We have $E_P = {p^2 \over 2M_P}$ and $E_Q = {p^2 \over 2M_Q}$. The total kinetic energy is $E_{total} = E_P + E_Q = \delta c^2$.

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

The term ${\mu = {{M_P M_Q} \over {M_P + M_Q}}}$ is the reduced mass. So, $\left({{M_Q + M_P} \over {M_P M_Q}}\right) = {1 \over \mu}$.

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

This matches option D. So, statement D is correct.

Conclusion:

• Option A is correct.
• Option B is incorrect.
• Option C is correct.
• Option D is correct.