A radioactive nucleus of mass number A, initially at rest, emits an `alpha-` particle with a speed v. What will be the recoil speed of the daughter nucleus ?

Options

A. `(A-4)/(4v)`
B. `(4v)/(A-4)`
C. v
D. `v/4`

Explanation

To solve the problem of finding the recoil speed of the daughter nucleus after a radioactive nucleus emits an alpha particle, we can use the principle of conservation of momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
Initially, we have a radioactive nucleus with mass number \( A \) at rest. When it emits an alpha particle (which has a mass number of 4), the remaining nucleus (daughter nucleus) will have a mass number of \( A - 4 \).

### Step 2: Write Down the 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.

- Initial momentum \( P_{\text{initial}} = 0 \) (since the nucleus is at rest).
- Final momentum \( P_{\text{final}} = P_{\text{daughter}} + P_{\text{alpha}} \).

### Step 3: Express the Final Momentum
Let \( v \) be the speed of the emitted alpha particle, and let \( v' \) be the recoil speed of the daughter nucleus. The mass of the alpha particle is \( 4 \) (in atomic mass units), and the mass of the daughter nucleus is \( A - 4 \).

Thus, we can express the final momentum as:
\[ P_{\text{final}} = (A - 4)v' + 4v \]

### Step 4: Set Up the Equation
Setting the initial momentum equal to the final momentum gives us:
\[ 0 = (A - 4)v' + 4v \]

### Step 5: Solve for the Recoil Speed \( v' \)
Rearranging the equation to solve for \( v' \):
\[ (A - 4)v' = -4v \]
\[ v' = -\frac{4v}{A - 4} \]

### Step 6: Interpret the Result
The negative sign indicates that the direction of the recoil speed of the daughter nucleus is opposite to the direction of the emitted alpha particle.

### Final Answer
The recoil speed of the daughter nucleus is given by:
\[ v' = -\frac{4v}{A - 4} \]