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, we will use the de Broglie wavelength formula, which relates the wavelength of a particle to its momentum. The formula is given by:

\[ \lambda = \frac{h}{P} \]

where:
- \(\lambda\) is the wavelength,
- \(h\) is the Planck constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(P\) is the momentum of the particle.

### Step-by-Step Solution:

1. **Identify the known values:**
- Wavelength of particle A, \(\lambda_A = 5 \times 10^{-8} \, \text{m}\).
- We need to find the wavelength of particle B, \(\lambda_B\), given that the momentum of B is double that of A.

2. **Express the momentum of particles A and B:**
- Let \(P_A\) be the momentum of particle A.
- According to the problem, the momentum of particle B is given as:
\[ P_B = 2P_A \]

3. **Write the de Broglie wavelength equations for both particles:**
- For particle A:
\[ \lambda_A = \frac{h}{P_A} \]
- For particle B:
\[ \lambda_B = \frac{h}{P_B} \]

4. **Substitute \(P_B\) in terms of \(P_A\) into the equation for \(\lambda_B\):**
- Since \(P_B = 2P_A\), we can write:
\[ \lambda_B = \frac{h}{2P_A} \]

5. **Relate \(\lambda_A\) and \(\lambda_B\):**
- We can express \(\lambda_B\) in terms of \(\lambda_A\):
\[ \lambda_B = \frac{h}{2P_A} = \frac{1}{2} \cdot \frac{h}{P_A} = \frac{1}{2} \lambda_A \]

6. **Calculate \(\lambda_B\):**
- Substitute the value of \(\lambda_A\):
\[ \lambda_B = \frac{1}{2} \cdot (5 \times 10^{-8} \, \text{m}) = 2.5 \times 10^{-8} \, \text{m} \]

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