A nucleus kept at rest in free space, brakes up into smaller nuclei of masses `'m'` and `'2m'`. Total energy generated in this fission is `E`. The bigger part is radioactive, emits five gamma ray photons in the direction opposite to its velocity and finally comes torest. `["Given" h=6.6xx10^(-34)Js,m=1xx10^(-26)Kg,E= 3.63xx10^(-8) mc^(2), C=3xx10^(8)m//s)`
Velocity of small daughter nucleus is

Options

A. `5.6 xx10^(4) m s^(-1)`
B. `6.6xx10^(4) m s^(-1)`
C. `7.6 xx 10^(4)m s^(-1)`
D. `8.6 xx 10^(4) m s^(-1)`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and conservation of energy.

### Step 1: Understand the System
We have a nucleus that breaks into two smaller nuclei with masses `m` and `2m`. The initial nucleus is at rest, meaning its initial momentum is zero. After the fission, the smaller nuclei will move in opposite directions.

### Step 2: Apply Conservation of Momentum
Since the initial momentum is zero, the momentum of the two smaller nuclei after the fission must also sum to zero:
\[ m v_1 + 2m v_2 = 0 \]
From this, we can express the velocities in terms of each other:
\[ v_1 = -2 v_2 \quad \text{(1)} \]
Here, \(v_1\) is the velocity of the smaller nucleus, and \(v_2\) is the velocity of the larger nucleus.

### Step 3: Apply Conservation of Energy
The total energy released during the fission is given as \(E\). The kinetic energy of both nuclei after the fission can be expressed as:
\[ E = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 \]
Substituting \(v_1\) from equation (1) into the energy equation:
\[ E = \frac{1}{2} m (-2v_2)^2 + \frac{1}{2} (2m) v_2^2 \]
This simplifies to:
\[ E = \frac{1}{2} m (4v_2^2) + \frac{1}{2} (2m) v_2^2 \]
\[ E = 2m v_2^2 + m v_2^2 = 3m v_2^2 \quad \text{(2)} \]

### Step 4: Substitute the Given Energy
We know from the problem statement that:
\[ E = 3.63 \times 10^{-8} mc^2 \]
Setting the two expressions for \(E\) equal gives:
\[ 3.63 \times 10^{-8} mc^2 = 3m v_2^2 \]
Dividing both sides by \(m\) (assuming \(m \neq 0\)):
\[ 3.63 \times 10^{-8} c^2 = 3 v_2^2 \]
Now, simplifying this:
\[ v_2^2 = \frac{3.63 \times 10^{-8} c^2}{3} = 1.21 \times 10^{-8} c^2 \]

### Step 5: Solve for \(v_2\)
Taking the square root of both sides:
\[ v_2 = c \sqrt{1.21 \times 10^{-8}} \]
Substituting \(c = 3 \times 10^8 \, \text{m/s}\):
\[ v_2 = 3 \times 10^8 \sqrt{1.21 \times 10^{-8}} = 3 \times 10^8 \times 1.1 \times 10^{-4} \]
Calculating this gives:
\[ v_2 \approx 3.3 \times 10^4 \, \text{m/s} \]

### Step 6: Find \(v_1\)
Using equation (1):
\[ v_1 = -2v_2 = -2(3.3 \times 10^4) \approx -6.6 \times 10^4 \, \text{m/s} \]
The negative sign indicates that \(v_1\) is in the opposite direction to \(v_2\).

### Final Answer
The velocity of the smaller daughter nucleus is:
\[ v_1 \approx 6.6 \times 10^4 \, \text{m/s} \]