If the velocity of the particle reduced to one third, then the percentage increase in its deBroglie wavelength is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of finding the percentage increase in the de Broglie wavelength when the velocity of a particle is reduced to one third, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Express the Wavelength in Terms of Velocity**:
Substituting the momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{mv} \]
This shows that the wavelength is inversely proportional to the velocity.

3. **Initial and Final Velocities**:
Let the initial velocity of the particle be \( v_1 \) and the final velocity after reduction be \( v_2 = \frac{1}{3} v_1 \).

4. **Relate the Initial and Final Wavelengths**:
Using the relationship derived from the de Broglie wavelength:
\[ \lambda_1 = \frac{h}{mv_1} \quad \text{and} \quad \lambda_2 = \frac{h}{mv_2} \]
Since \( v_2 = \frac{1}{3} v_1 \), we can substitute this into the equation for \( \lambda_2 \):
\[ \lambda_2 = \frac{h}{m(\frac{1}{3}v_1)} = \frac{3h}{mv_1} = 3\lambda_1 \]

5. **Calculate the Change in Wavelength**:
The change in wavelength (\( \Delta \lambda \)) is given by:
\[ \Delta \lambda = \lambda_2 - \lambda_1 = 3\lambda_1 - \lambda_1 = 2\lambda_1 \]

6. **Calculate the Percentage Increase**:
The percentage increase in wavelength is calculated as:
\[ \text{Percentage Increase} = \left(\frac{\Delta \lambda}{\lambda_1}\right) \times 100 = \left(\frac{2\lambda_1}{\lambda_1}\right) \times 100 = 200\% \]

### Final Answer:
The percentage increase in the de Broglie wavelength when the velocity of the particle is reduced to one third is **200%**.