Two particles A and B are in motion if wavelength of A is `5 xx 10^(-8)m`, calculate wavelength of B so that its momentum is double of A

Options

A. `5 xx 10^(-8)m`
B. `2.5 xx 10^(-8)m`
C. `4 xx 10^(-8) m`
D. `3.5 xx 10^(-8)m`

Answer

To solve the problem of finding the wavelength of particle B given that its momentum is double that of particle A, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Wavelength of particle A, \( \lambda_A = 5 \times 10^{-8} \, m \)
- Momentum of particle B, \( P_B = 2 \times P_A \)

2. **Use de Broglie's Wavelength Formula**:
According to de Broglie's hypothesis, the wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

3. **Write the Momentum for Particles A and B**:
For particle A:
\[ \lambda_A = \frac{h}{P_A} \]
For particle B:
\[ \lambda_B = \frac{h}{P_B} \]

4. **Substitute the Momentum of B**:
Since \( P_B = 2 \times P_A \), we can write:
\[ \lambda_B = \frac{h}{2 \times P_A} \]

5. **Relate Wavelengths of A and B**:
From the equation for \( \lambda_A \):
\[ P_A = \frac{h}{\lambda_A} \]
Substitute this into the equation for \( \lambda_B \):
\[ \lambda_B = \frac{h}{2 \times \frac{h}{\lambda_A}} = \frac{\lambda_A}{2} \]

6. **Calculate Wavelength of B**:
Substitute the value of \( \lambda_A \):
\[ \lambda_B = \frac{5 \times 10^{-8} \, m}{2} = 2.5 \times 10^{-8} \, m \]

7. **Final Answer**:
The wavelength of particle B is:
\[ \lambda_B = 2.5 \times 10^{-8} \, m \]

### Summary:
The wavelength of particle B, given that its momentum is double that of particle A, is \( 2.5 \times 10^{-8} \, m \).