A nucleus of `._84^210 Po` originally at rest emits `alpha` particle with speed `v`. What will be the recoil speed of the daughter nucleus ?

Options

A. `4v//206`
B. `4v//214`
C. `v//206`
D. `v//214`

Answer

To solve the problem of finding the recoil speed of the daughter nucleus after the emission of an alpha particle from a polonium-210 nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The nucleus of polonium-210 (denoted as \( _{84}^{210}Po \)) is initially at rest. Therefore, the initial momentum of the system is zero.

2. **Define the Masses**:
- The mass of the emitted alpha particle is approximately 4 units (since an alpha particle is essentially a helium nucleus).
- The mass of the daughter nucleus after the emission of the alpha particle can be calculated as:
\[ \text{Mass of daughter nucleus} = 210 - 4 = 206 \text{ units} \]

3. **Apply Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum before the emission must equal the total momentum after the emission.
- Initially, the momentum is zero:
\[ \text{Initial Momentum} = 0 \]
- After the emission, the momentum can be expressed as:
\[ \text{Final Momentum} = \text{Momentum of alpha particle} + \text{Momentum of daughter nucleus} \]
- Let \( v \) be the speed of the emitted alpha particle and \( V_1 \) be the recoil speed of the daughter nucleus. The momentum of the alpha particle is \( 4v \) (in the negative direction) and the momentum of the daughter nucleus is \( 206V_1 \) (in the positive direction).

4. **Set Up the Equation**:
- Since the initial momentum is zero, we can set up the equation:
\[ 0 = -4v + 206V_1 \]

5. **Solve for the Recoil Speed \( V_1 \)**:
- Rearranging the equation gives:
\[ 206V_1 = 4v \]
- Dividing both sides by 206, we find:
\[ V_1 = \frac{4v}{206} \]

6. **Final Result**:
- The recoil speed of the daughter nucleus is:
\[ V_1 = \frac{4v}{206} \]

### Summary:
The recoil speed of the daughter nucleus after the emission of the alpha particle is \( \frac{4v}{206} \).