A stationary shell of mass M explodes in to two parts and their masses are in the ratio 2:3, then the ratio of their deBroglie wavelengths is

Options

A. `2:3`
B. `3:2`
C. `9:4`
D. `1:1`

Answer

To solve the problem of finding the ratio of the de Broglie wavelengths of two parts after a stationary shell of mass M explodes into two parts with masses in the ratio 2:3, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses**:
Let the masses of the two parts be \( m_1 \) and \( m_2 \).
Given the ratio of their masses is \( 2:3 \), we can express the masses as:
\[ m_1 = \frac{2}{5}M \quad \text{and} \quad m_2 = \frac{3}{5}M \]
where \( M \) is the total mass of the shell.

2. **Conservation of Momentum**:
Since the shell is initially at rest, the total momentum before the explosion is zero. After the explosion, the two parts will move in opposite directions, and their momenta must cancel each other out:
\[ P_1 + P_2 = 0 \]
This implies:
\[ P_1 = -P_2 \]

3. **Expressing Momentum**:
The momentum of each part can be expressed as:
\[ P_1 = m_1 v_1 \quad \text{and} \quad P_2 = m_2 v_2 \]
Therefore, we have:
\[ m_1 v_1 = m_2 v_2 \]

4. **Finding the Velocities**:
Rearranging the momentum equation gives us:
\[ v_1 = \frac{m_2}{m_1} v_2 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ v_1 = \frac{\frac{3}{5}M}{\frac{2}{5}M} v_2 = \frac{3}{2} v_2 \]

5. **Calculating de Broglie Wavelengths**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum. Thus, we can write:
\[ \lambda_1 = \frac{h}{m_1 v_1} \quad \text{and} \quad \lambda_2 = \frac{h}{m_2 v_2} \]

6. **Substituting for Velocities**:
Substitute \( v_1 \) into the expression for \( \lambda_1 \):
\[ \lambda_1 = \frac{h}{m_1 \left(\frac{3}{2} v_2\right)} = \frac{2h}{3m_1 v_2} \]
Now, substituting \( m_1 = \frac{2}{5}M \):
\[ \lambda_1 = \frac{2h}{3 \left(\frac{2}{5}M\right) v_2} = \frac{5h}{3Mv_2} \]

7. **Finding the Ratio of Wavelengths**:
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{5h}{3Mv_2}}{\frac{h}{m_2 v_2}} = \frac{5}{3} \cdot \frac{m_2}{M} \]
Substituting \( m_2 = \frac{3}{5}M \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{5}{3} \cdot \frac{\frac{3}{5}M}{M} = \frac{5}{3} \cdot \frac{3}{5} = 1 \]

8. **Final Result**:
Therefore, the ratio of their de Broglie wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = 1:1 \]