Question

A heavy nucleus $N$, at rest, undergoes fission $N \rightarrow P + Q$, 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 speeds 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?

• A. $E_{p}+E_{Q}=c^{2} \delta$
• B. $E_{p}=\left(\frac{M_{p}}{M_{p}+M_{Q}}\right) c^{2} \delta$
• 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 $\mu=\frac{ M _{ p } M _{ Q }}{\left( M _{ p }+ M _{ Q }\right)}$

Answer 1

The Correct Option is A, C, D

Step 1: Given Information
A heavy nucleus \( N \), at rest, undergoes fission as: \[ N \rightarrow P + Q \] where \( P \) and \( Q \) are two lighter nuclei. We are given that: - \( \delta = M_N - M_P - M_Q \), where \( M_N, M_P, M_Q \) are the masses of the nuclei \( N, P, Q \), respectively.
- \( E_P \) and \( E_Q \) are the kinetic energies of \( P \) and \( Q \), respectively.
- The speeds of \( P \) and \( Q \) are \( v_P \) and \( v_Q \), respectively.
- \( c \) is the speed of light.
We need to verify the correctness of the following statements: - (A) \( E_P + E_Q = c^2 \delta \) - (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 \( \mu = \frac{M_P M_Q}{M_P + M_Q} \)

Step 2: Energy Considerations
From the principle of conservation of energy, the total energy before and after the fission must be conserved. Initially, the nucleus \( N \) is at rest, so its total energy is simply its rest mass energy: \[ E_{\text{initial}} = M_N c^2 \] After the fission, the total energy of the system is the sum of the rest mass energies of \( P \) and \( Q \) and their kinetic energies: \[ E_{\text{final}} = M_P c^2 + M_Q c^2 + E_P + E_Q \] By conservation of energy, we have: \[ M_N c^2 = M_P c^2 + M_Q c^2 + E_P + E_Q \] Simplifying this equation: \[ E_P + E_Q = (M_N - M_P - M_Q) c^2 = \delta c^2 \] Therefore, statement (A) is correct: \[ E_P + E_Q = c^2 \delta \]

Step 3: Momentum Considerations
Since the total momentum of the system before the fission is zero (because \( N \) is initially at rest), the total momentum after the fission must also be zero. Therefore, the momentum of \( P \) and \( Q \) must be equal in magnitude but opposite in direction: \[ M_P v_P = M_Q v_Q \] This implies: \[ \frac{v_P}{v_Q} = \frac{M_Q}{M_P} \] Therefore, statement (C) is correct: \[ \frac{v_P}{v_Q} = \frac{M_Q}{M_P} \]

Step 4: Magnitude of Momentum
The total momentum of the system must be zero, so the magnitudes of the momenta of \( P \) and \( Q \) are equal. The relativistic momentum of each particle is given by: \[ p_P = M_P v_P \quad \text{and} \quad p_Q = M_Q v_Q \] Since the magnitudes of the momenta are equal: \[ M_P v_P = M_Q v_Q \] Using the relation \( \frac{v_P}{v_Q} = \frac{M_Q}{M_P} \), we find: \[ p_P = p_Q = \sqrt{2 \mu \delta} \] where \( \mu = \frac{M_P M_Q}{M_P + M_Q} \) is the reduced mass of the system. The magnitude of the momentum for both \( P \) and \( Q \) is therefore \( c \sqrt{2 \mu \delta} \), so statement (D) is correct: \[ p_P = p_Q = c \sqrt{2 \mu \delta} \]

Final Answer:
All the statements (A), (C), and (D) are correct:
- (A) \( E_P + E_Q = c^2 \delta \)
- (C) \( \frac{v_P}{v_Q} = \frac{M_Q}{M_P} \)
- (D) The magnitude of momentum for \( P \) as well as \( Q \) is \( c \sqrt{2 \mu \delta} \)