After absorbing a slowly moving neutrons of mass `m_(N)` (momentum ~0) a nucleus of mass M breaks into two nucleii of mass `m_(1) and 5m_(1)(6m_(1)=M+m_(N))`, respectively . If the de-Broglie wavelength of the nucleus with mass `m_(1) ` is ` lambda`, then de Broglie wavelength of the other nucleus will be

Options

A. `25lambda`
B. `5lambda`
C. `(lambda)/(5)`
D. `lambda`

Answer

To solve the problem, we need to analyze the situation using the principles of conservation of momentum and the de Broglie wavelength formula.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We have a nucleus of mass \( M \) that absorbs a neutron of mass \( m_N \) and breaks into two smaller nuclei of masses \( m_1 \) and \( 5m_1 \). The total mass before the reaction is equal to the total mass after the reaction.

2. **Setting Up the Mass Equation:**
From the problem, we know:
\[ M + m_N = 6m_1 \]
This means that the total mass of the system is conserved.

3. **Conservation of Momentum:**
Since the neutron is slowly moving, we can assume its momentum is negligible (approximately zero). Therefore, the initial momentum \( p_{\text{initial}} \) is:
\[ p_{\text{initial}} = 0 \]
After the nucleus breaks into two parts, the momentum conservation gives us:
\[ p_{\text{initial}} = p_1 + p_2 = 0 \]
This implies:
\[ p_1 = -p_2 \]

4. **Relating Momentum to Mass and Velocity:**
The momentum of each nucleus can be expressed as:
\[ p_1 = m_1 v_1 \quad \text{and} \quad p_2 = 5m_1 v_2 \]
From the conservation of momentum, we have:
\[ m_1 v_1 = 5m_1 v_2 \]
Dividing both sides by \( m_1 \) (assuming \( m_1 \neq 0 \)):
\[ v_1 = 5 v_2 \]

5. **Calculating the de Broglie Wavelength:**
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum. For nucleus with mass \( m_1 \):
\[ \lambda_1 = \frac{h}{p_1} = \frac{h}{m_1 v_1} \]
For nucleus with mass \( 5m_1 \):
\[ \lambda_2 = \frac{h}{p_2} = \frac{h}{5m_1 v_2} \]

6. **Relating the Wavelengths:**
Since \( v_1 = 5 v_2 \), we can substitute \( v_1 \) in the equation for \( \lambda_1 \):
\[ \lambda_1 = \frac{h}{m_1 (5 v_2)} = \frac{1}{5} \cdot \frac{h}{m_1 v_2} = \frac{1}{5} \lambda_2 \]
Rearranging gives:
\[ \lambda_2 = 5 \lambda_1 \]

7. **Final Result:**
If the de Broglie wavelength of the nucleus with mass \( m_1 \) is \( \lambda \), then the de Broglie wavelength of the other nucleus (mass \( 5m_1 \)) will be:
\[ \lambda_2 = 5\lambda \]