If mass of an atom is M moving with speed v, what will be its speed after the emission of an `alpha`-particle if speed of `alpha`-particle is zero ?

Options

A. (a) `(Mv)/(M+2)`
B. (b) `(Mv)/(M-4)`
C. (c) `(Mv)/(M+4)`
D. (d) `(M-4)/(Mv)`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The mass of the atom is \( M \).
- The initial speed of the atom is \( v \).
- The alpha particle has a mass of 4 (let's denote it as \( m_{\alpha} = 4 \)).
- The speed of the alpha particle after emission is given as \( 0 \).

2. **Calculate the Initial Momentum:**
- The initial momentum (\( p_{\text{initial}} \)) of the system (the atom) can be calculated as:
\[ p_{\text{initial}} = M \cdot v \]

3. **Determine the Final Conditions:**
- After the emission of the alpha particle, the remaining mass of the atom will be \( M - 4 \).
- Let the final speed of the remaining atom be \( x \).

4. **Calculate the Final Momentum:**
- The final momentum (\( p_{\text{final}} \)) of the system (the remaining atom) can be calculated as:
\[ p_{\text{final}} = (M - 4) \cdot x \]

5. **Apply Conservation of Momentum:**
- According to the conservation of momentum, the initial momentum must equal the final momentum:
\[ M \cdot v = (M - 4) \cdot x \]

6. **Solve for the Final Speed \( x \):**
- Rearranging the equation to solve for \( x \):
\[ x = \frac{M \cdot v}{M - 4} \]

### Final Answer:
The speed of the atom after the emission of the alpha particle is:
\[ x = \frac{M \cdot v}{M - 4} \]