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 D

The problem involves the fission of a heavy nucleus \( N \), which undergoes the reaction:

\( N \rightarrow P + Q \)

Where \( P \) and \( Q \) are two lighter nuclei formed during the fission process. The masses of the nuclei are denoted by \( M_N \), \( M_P \), and \( M_Q \) for \( N \), \( P \), and \( Q \), respectively. The quantity \( \delta \) is defined as:

\( \delta = M_N - M_P - M_Q \)

Here, \( E_P \) and \( E_Q \) are the kinetic energies of the nuclei \( P \) and \( Q \), respectively. The speeds of \( P \) and \( Q \) are given as \( v_P \) and \( v_Q \), and \( c \) is the speed of light.

Step 1: Conservation of Energy

In the fission process, the total energy is conserved. The initial rest mass energy of the nucleus \( N \) is given by \( E_N = M_N c^2 \). After the fission, the total energy is the sum of the rest mass energies and the kinetic energies of the fission products \( P \) and \( Q \):

\( E_{\text{total}} = M_P c^2 + M_Q c^2 + E_P + E_Q \)

Using the principle of conservation of energy:

\( M_N c^2 = M_P c^2 + M_Q c^2 + E_P + E_Q \)

This implies that the sum of the kinetic energies \( E_P + E_Q \) is related to the mass defect \( \delta \) by:

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

Step 2: Conservation of Momentum

Since the nucleus \( N \) is initially at rest, the total momentum of the system must be zero after the fission. This means that the momenta of \( P \) and \( Q \) must be equal in magnitude and opposite in direction:

\( \vec{p}_P = - \vec{p}_Q \)

The magnitudes of the momenta are given by:

\( p_P = p_Q \)

Step 3: Relating Momentum to Kinetic Energy

The kinetic energy of each particle is related to its momentum by the relativistic equation:

\( E = \sqrt{p^2 c^2 + M^2 c^4} - M c^2 \)

For \( P \), the energy is:

\( E_P = \sqrt{p_P^2 c^2 + M_P^2 c^4} - M_P c^2 \)

Similarly, for \( Q \), the energy is:

\( E_Q = \sqrt{p_Q^2 c^2 + M_Q^2 c^4} - M_Q c^2 \)

Step 4: Result for Momentum

Since \( p_P = p_Q \), the magnitude of the momentum for both particles is the same. By using the conservation of energy and momentum, it can be shown that the magnitude of the momentum for both \( P \) and \( Q \) is given by:

\( p_P = p_Q = c \sqrt{2 \mu \delta} \)

Where \( \mu \) is the reduced mass, defined as:

\( \mu = \frac{M_P M_Q}{M_P + M_Q} \)

Final Answer:

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