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. `v/210`
B. `v/84`
C. `"4v"/206`
D. `"3v"/214`

Answer

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

### Step 1: Understand the Problem
We have a polonium nucleus \( _{84}^{210}Po \) that is initially at rest. It emits an alpha particle (which is a helium nucleus \( _{2}^{4}He \)) with a speed \( v \). We need to find the recoil speed of the daughter nucleus, which is the remaining nucleus after the alpha particle is emitted.

### Step 2: Write the Reaction
When the polonium nucleus emits an alpha particle, the reaction can be written as:
\[ _{84}^{210}Po \rightarrow _{2}^{4}He + X \]
where \( X \) is the daughter nucleus. The mass number of the daughter nucleus \( X \) can be calculated as:
\[ A_X = 210 - 4 = 206 \]
and the atomic number will be:
\[ Z_X = 84 - 2 = 82 \]
Thus, the daughter nucleus is \( _{82}^{206}Pb \).

### Step 3: Apply Conservation of Momentum
Since the initial momentum of the system is zero (the nucleus is at rest), the final momentum must also be zero. Let \( v' \) be the recoil speed of the daughter nucleus. The conservation of momentum can be expressed as:
\[ 0 = m_{He} v + m_{Pb} v' \]
where \( m_{He} \) is the mass of the alpha particle and \( m_{Pb} \) is the mass of the daughter nucleus.

### Step 4: Substitute Masses
The masses can be approximated based on the mass numbers:
- Mass of alpha particle \( m_{He} \approx 4 \) (for \( _{2}^{4}He \))
- Mass of daughter nucleus \( m_{Pb} \approx 206 \) (for \( _{82}^{206}Pb \))

Substituting these values into the momentum equation gives:
\[ 0 = 4v + 206v' \]

### Step 5: Solve for Recoil Speed
Rearranging the equation to solve for \( v' \):
\[ 206v' = -4v \]
\[ v' = -\frac{4v}{206} \]
Taking the magnitude (since speed is a scalar quantity):
\[ |v'| = \frac{4v}{206} \]

### Step 6: Final Answer
Thus, the recoil speed of the daughter nucleus is:
\[ v' = \frac{4v}{206} \]

### Conclusion
The correct option from the provided choices is option 3: \( \frac{4v}{206} \).

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