A : If a stationary nucleus emits an ` alpha` -perticle , the de Broglie wavelengths of te daugther nucleus and the ` alpha`-perticle and equal .
R : The magnitudes of the linear moments of the daughter nucleus and the `alpha`-perticle are the same .

Options

A. If both Assertion & Reason are ture and the reason is the correct explanation of the assertion , then mark (1) .
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion , then mark (2) .
C. If Assertion is true statement but Reason is false , then mark (3)
D. If both Assertion and Reason are false statements , then mark (4) .

Answer

To solve the question, we need to analyze the assertion (A) and reason (R) provided:

**Assertion (A):** If a stationary nucleus emits an alpha particle, the de Broglie wavelengths of the daughter nucleus and the alpha particle are equal.

**Reason (R):** The magnitudes of the linear momenta of the daughter nucleus and the alpha particle are the same.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- We start with a stationary nucleus, which means both the nucleus and the alpha particle have an initial momentum of zero.
- Let’s denote the daughter nucleus as \( N' \) and the emitted alpha particle as \( \alpha \).

2. **Applying Conservation of Linear 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:
\[ p_{\text{initial}} = 0 \]
- After the emission, the momentum can be expressed as:
\[ p_{\text{final}} = p_{\alpha} + p_{N'} \]
- Setting the initial and final momentum equal gives us:
\[ 0 = p_{\alpha} + p_{N'} \]
- This implies:
\[ p_{\alpha} = -p_{N'} \]
- Therefore, the magnitudes of the momenta are equal:
\[ |p_{\alpha}| = |p_{N'}| \]

3. **Relating Momentum to de Broglie Wavelength:**
- The de Broglie wavelength \( \lambda \) is related to momentum \( p \) by the formula:
\[ \lambda = \frac{h}{p} \]
- For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{|p_{\alpha}|} \]
- For the daughter nucleus:
\[ \lambda_{N'} = \frac{h}{|p_{N'}|} \]
- Since \( |p_{\alpha}| = |p_{N'}| \), we can conclude:
\[ \lambda_{\alpha} = \lambda_{N'} \]

4. **Conclusion:**
- Since we have shown that the de Broglie wavelengths of the daughter nucleus and the alpha particle are equal, the assertion (A) is correct.
- Furthermore, since we established that the magnitudes of their momenta are the same, the reason (R) is also correct.
- Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

### Final Answer:
Both assertion (A) and reason (R) are correct, and the reason provides the correct explanation for the assertion.

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