Nuclear Physics

The problem involves the fission of a heavy nucleus , which undergoes the reaction:

Where and are two lighter nuclei formed during the fission process. The masses of the nuclei are denoted by , , and for , , and , respectively. The quantity is defined as:

Here, and are the kinetic energies of the nuclei and , respectively. The speeds of and are given as and , and 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 is given by . After the fission, the total energy is the sum of the rest mass energies and the kinetic energies of the fission products and :

Using the principle of conservation of energy:

This implies that the sum of the kinetic energies is related to the mass defect by:

Step 2: Conservation of Momentum

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

The magnitudes of the momenta are given by:

Step 3: Relating Momentum to Kinetic Energy

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

For , the energy is:

Similarly, for , the energy is:

Step 4: Result for Momentum

Since , 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 and is given by:

Where is the reduced mass, defined as:

Final Answer:

The magnitude of momentum for as well as is , where .

Step 1: Given Information
A heavy nucleus , at rest, undergoes fission as: where and are two lighter nuclei. We are given that: - , where are the masses of the nuclei , respectively.
- and are the kinetic energies of and , respectively.
- The speeds of and are and , respectively.
- is the speed of light.
We need to verify the correctness of the following statements: - (A) - (C) - (D) The magnitude of momentum for as well as is , where

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 is at rest, so its total energy is simply its rest mass energy: After the fission, the total energy of the system is the sum of the rest mass energies of and and their kinetic energies: By conservation of energy, we have: Simplifying this equation: Therefore, statement (A) is correct:

Step 3: Momentum Considerations
Since the total momentum of the system before the fission is zero (because is initially at rest), the total momentum after the fission must also be zero. Therefore, the momentum of and must be equal in magnitude but opposite in direction: This implies: Therefore, statement (C) is correct:

Step 4: Magnitude of Momentum
The total momentum of the system must be zero, so the magnitudes of the momenta of and are equal. The relativistic momentum of each particle is given by: Since the magnitudes of the momenta are equal: Using the relation , we find: where is the reduced mass of the system. The magnitude of the momentum for both and is therefore , so statement (D) is correct:

Final Answer:
All the statements (A), (C), and (D) are correct:
- (A)
- (C)
- (D) The magnitude of momentum for as well as is

is proportional to where is the atomic number of the nucleus,
is proportional to where is the mass number of the nucleus and
increases if the nucleus undergoes a beta decay emitting a positron.

The number of radioactive nuclei N(t) at a given time t is given by:
N(t) = N₀ exp(-λ_t)
The activity A of a radioactive source is defined as the rate of decay, which is the derivative of the number of radioactive nuclei with respect to time: A = - = λ N(t)
Given that the activities of sources S₁ and S₂ are A₁ and A₂, respectively, we can write:
A₁ = λ₁ N₁(t)
A₂ = λ₂ N₂(t)
After the completion of the 3^rd half-life for source S₁, the remaining number of radioactive nuclei is:
N₁(t) = N₀ ( )³ = N₀
Similarly, after the completion of the 7^th half-life for source S₂, the remaining number of radioactive nuclei is:
N₂(t) = N₀ ⁷ = N₀
Now, let's substitute these values into the equations for activities A₁ and A₂ :
A₁ = λ₁ N₀ ( )

A₂ = λ₂ N₀ ( )

To find the ratio , we can simplify:

( ) = ( ) ( ) ( )
Since both sources have the same activity at time t = 0, we can assume their decay constants are the same (λ₁ = λ₂). Therefore, the ratio simplifies to:
( ) = = = 16

Therefore, the ratio is 16.

So, the answer is .