Two particles A and B are in motion. If the wavelength associated with particle A is `5 xx 10^(-8)m`, calculate the wavelength associated with particle B if its momentum is half of A.

Options

A. `5xx 10 ^(-8)m`
B. `10 ^(-5) cm`
C. `10^(-7)`cm
D. `5xx 10^(-8) cm`

Answer

To solve the problem, we will use the de Broglie wavelength formula, which relates the wavelength (λ) of a particle to its momentum (P). The formula is given by:

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

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

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Wavelength of particle A (\( \lambda_A \)) = \( 5 \times 10^{-8} \, \text{m} \)
- Momentum of particle B (\( P_B \)) is half of momentum of particle A (\( P_A \)).
- Therefore, \( P_B = \frac{1}{2} P_A \).

2. **Express the Relationship Between Wavelengths and Momenta**:
- According to the de Broglie equation, we can write:
\[ \frac{\lambda_A}{\lambda_B} = \frac{P_B}{P_A} \]

3. **Substituting the Known Values**:
- Since \( P_B = \frac{1}{2} P_A \), we can substitute this into the equation:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\frac{1}{2} P_A}{P_A} = \frac{1}{2} \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ \lambda_B = 2 \lambda_A \]

5. **Calculating Wavelength of Particle B**:
- Substitute \( \lambda_A = 5 \times 10^{-8} \, \text{m} \):
\[ \lambda_B = 2 \times (5 \times 10^{-8} \, \text{m}) = 10 \times 10^{-8} \, \text{m} = 1 \times 10^{-7} \, \text{m} \]

6. **Convert to Centimeters**:
- To convert meters to centimeters (1 m = 100 cm):
\[ \lambda_B = 1 \times 10^{-7} \, \text{m} \times 100 \, \text{cm/m} = 1 \times 10^{-5} \, \text{cm} \]

### Final Answer:
The wavelength associated with particle B is \( \lambda_B = 1 \times 10^{-5} \, \text{cm} \).

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