A radioactive nucleus undergoes `alpha-`emission to form a stable element. What will be the recoil velocity of the daughter nucleus is `V` is the velocity of `alpha`-emission and `A` is the atomic mass of radioactive nucleus ?

Options

A. `(4V)/(A-4)`
B. `(2V)/(A-4)`
C. `(4V)/(A+4)`
D. `(2V)/(A+4)`

Answer

To solve the problem of finding the recoil velocity of the daughter nucleus after alpha emission, we can follow these steps:

### Step 1: Understand the Reaction
In alpha emission, a radioactive nucleus (let's denote its mass as \( A \)) emits an alpha particle (which has a mass of 4 units) and transforms into a daughter nucleus. The mass of the daughter nucleus will therefore be \( A - 4 \).

### Step 2: Apply Conservation of Momentum
Since momentum is conserved during the emission process, we can set up the equation based on the initial and final momentum of the system. Initially, the radioactive nucleus is at rest, so the total initial momentum is zero.

### Step 3: Define the Variables
Let:
- \( V \) = velocity of the emitted alpha particle
- \( V_1 \) = recoil velocity of the daughter nucleus
- Mass of the alpha particle = 4
- Mass of the daughter nucleus = \( A - 4 \)

### Step 4: Write the Momentum Conservation Equation
The total momentum after the emission must equal the total momentum before the emission. Therefore, we can write:

\[ 0 = (4)(V) + (A - 4)(-V_1) \]

### Step 5: Rearrange the Equation
Rearranging the equation gives us:

\[ (4)(V) = (A - 4)(V_1) \]

### Step 6: Solve for \( V_1 \)
Now, we can solve for the recoil velocity \( V_1 \):

\[ V_1 = \frac{4V}{A - 4} \]

### Conclusion
Thus, the recoil velocity of the daughter nucleus after the alpha emission is given by:

\[ V_1 = \frac{4V}{A - 4} \]