A normal stationary lithium atom releases an `alpha`-particle with velocity `(2hati+3hatj+7hatk)`km/s .
What will be the velocity of the daughter element just after the releases of the `alpha`-particle ?

Options

A. `(1//2)(2hati+3hatj+7hatk)`km/s
B. `(-4//3)(2hati+3hatj+7hatk)`km/s
C. `(-1//4) (2hati+3hatj+7hatk)` km/s
D. `(1//5)(2hati+3hatj+7hatk)` km/s

Answer

To solve the problem, we will use the principle of conservation of linear momentum. Here are the steps to find the velocity of the daughter element after the alpha particle is released from the lithium atom.

### Step 1: Understand the system
A lithium atom (Li) at rest releases an alpha particle (α). The lithium atom has an atomic number of 3 and a mass number of 7. The alpha particle has an atomic number of 2 and a mass number of 4. After the release, the lithium atom transforms into a tritium atom (T), which has an atomic number of 1 and a mass number of 3.

### Step 2: Write down the initial momentum
Since the lithium atom is initially at rest, its initial momentum is zero. Therefore, the total initial momentum of the system (lithium atom + alpha particle) is:
\[ \text{Initial Momentum} = 0 \]

### Step 3: Write down the final momentum
After the alpha particle is released, the final momentum of the system can be expressed as the sum of the momentum of the alpha particle and the momentum of the daughter element (tritium):
\[ \text{Final Momentum} = \text{Momentum of α} + \text{Momentum of T} \]

### Step 4: Calculate the momentum of the alpha particle
The momentum of the alpha particle is given by:
\[ \text{Momentum of α} = m_{\alpha} \cdot \vec{v}_{\alpha} \]
where \( m_{\alpha} = 4 \) (mass of the alpha particle) and \( \vec{v}_{\alpha} = (2 \hat{i} + 3 \hat{j} + 7 \hat{k}) \) km/s.

Thus,
\[ \text{Momentum of α} = 4 \cdot (2 \hat{i} + 3 \hat{j} + 7 \hat{k}) = (8 \hat{i} + 12 \hat{j} + 28 \hat{k}) \text{ kg km/s} \]

### Step 5: Set up the equation for conservation of momentum
According to the conservation of momentum:
\[ 0 = \text{Momentum of α} + \text{Momentum of T} \]
Let \( \vec{v}_T \) be the velocity of the tritium. The momentum of the tritium is:
\[ \text{Momentum of T} = m_T \cdot \vec{v}_T \]
where \( m_T = 3 \) (mass of tritium).

So, we have:
\[ 0 = (8 \hat{i} + 12 \hat{j} + 28 \hat{k}) + 3 \cdot \vec{v}_T \]

### Step 6: Solve for the velocity of the daughter element
Rearranging the equation gives:
\[ 3 \cdot \vec{v}_T = - (8 \hat{i} + 12 \hat{j} + 28 \hat{k}) \]
\[ \vec{v}_T = -\frac{1}{3} (8 \hat{i} + 12 \hat{j} + 28 \hat{k}) \]
\[ \vec{v}_T = -\left(\frac{8}{3} \hat{i} + 4 \hat{j} + \frac{28}{3} \hat{k}\right) \text{ km/s} \]

### Final Answer
The velocity of the daughter element (tritium) just after the release of the alpha particle is:
\[ \vec{v}_T = -\left(\frac{8}{3} \hat{i} + 4 \hat{j} + \frac{28}{3} \hat{k}\right) \text{ km/s} \]