A neutron of mass `1.67xx10^(-27)` kg moving with a speed of `3xx10^(6)ms^(-1)` collides with a deutron of mass `3.34xx10^(-27)` kg which is at rest. After collision, the neutron sticks to the deutron and forms a triton. What is the speed of the triton?

Options

A. `2xx10^(6)ms^(-1)`
B. `4xx10^(6)ms^(-1)`
C. `6xx10^(6)ms^(-1)`
D. `8xx10^(6)ms^(-1)`

Answer

To solve the problem of a neutron colliding with a deuteron and forming a triton, we will use the principle of conservation of linear momentum. Here’s a step-by-step solution:

### Step 1: Identify the Given Values
- Mass of the neutron (m1) = \(1.67 \times 10^{-27}\) kg
- Initial speed of the neutron (u1) = \(3 \times 10^{6}\) m/s
- Mass of the deuteron (m2) = \(3.34 \times 10^{-27}\) kg
- Initial speed of the deuteron (u2) = 0 m/s (since it is at rest)

### Step 2: Write the Conservation of Momentum Equation
According to the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
Where:
- \(v\) is the final speed of the triton formed after the collision.

### Step 3: Substitute the Known Values
Substituting the known values into the equation:
\[ (1.67 \times 10^{-27} \, \text{kg}) \times (3 \times 10^{6} \, \text{m/s}) + (3.34 \times 10^{-27} \, \text{kg}) \times (0) = (1.67 \times 10^{-27} \, \text{kg} + 3.34 \times 10^{-27} \, \text{kg}) v \]
This simplifies to:
\[ (1.67 \times 10^{-27} \times 3 \times 10^{6}) = (5.01 \times 10^{-27}) v \]

### Step 4: Calculate the Left Side
Calculating the left side:
\[ 1.67 \times 3 = 5.01 \]
Thus:
\[ 5.01 \times 10^{-21} = (5.01 \times 10^{-27}) v \]

### Step 5: Solve for v
Now, we can solve for \(v\):
\[ v = \frac{5.01 \times 10^{-21}}{5.01 \times 10^{-27}} \]
This simplifies to:
\[ v = 10^{6} \, \text{m/s} \]

### Final Answer
The speed of the triton after the collision is:
\[ v = 10^{6} \, \text{m/s} \]