If the kinetic energy of a particle is increased by 10 times, the percentage change in the de Broglie wavelength of the particle is

Options

A. `25%`
B. `75%`
C. `60%`
D. `50%`

Explanation

To solve the problem of finding the percentage change in the de Broglie wavelength of a particle when its kinetic energy is increased by 10 times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie wavelength formula**:
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Relate kinetic energy to velocity**:
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express the velocity in terms of kinetic energy:
\[ v = \sqrt{\frac{2KE}{m}} \]

3. **Substitute velocity into the de Broglie wavelength formula**:
Substituting the expression for \( v \) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m \sqrt{\frac{2KE}{m}}} = \frac{h}{\sqrt{2m \cdot KE}} \]

4. **Determine the initial and final kinetic energies**:
Let the initial kinetic energy be \( KE_1 \). If the kinetic energy is increased by 10 times, the new kinetic energy \( KE_2 \) is:
\[ KE_2 = 10 \cdot KE_1 \]

5. **Calculate the initial and final wavelengths**:
Using the de Broglie wavelength formula:
- For the initial wavelength:
\[ \lambda_1 = \frac{h}{\sqrt{2m \cdot KE_1}} \]
- For the final wavelength:
\[ \lambda_2 = \frac{h}{\sqrt{2m \cdot KE_2}} = \frac{h}{\sqrt{2m \cdot (10 \cdot KE_1)}} = \frac{h}{\sqrt{10} \cdot \sqrt{2m \cdot KE_1}} = \frac{\lambda_1}{\sqrt{10}} \]

6. **Find the percentage change in wavelength**:
The percentage change in wavelength can be calculated using the formula:
\[ \text{Percentage Change} = \frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100 \]
Substituting \( \lambda_2 = \frac{\lambda_1}{\sqrt{10}} \):
\[ \text{Percentage Change} = \frac{\lambda_1 - \frac{\lambda_1}{\sqrt{10}}}{\lambda_1} \times 100 = \left(1 - \frac{1}{\sqrt{10}}\right) \times 100 \]

7. **Calculate the numerical value**:
Now, calculate \( \frac{1}{\sqrt{10}} \):
\[ \sqrt{10} \approx 3.162 \implies \frac{1}{\sqrt{10}} \approx 0.316 \]
Thus:
\[ \text{Percentage Change} = (1 - 0.316) \times 100 \approx 68.4\% \]

### Final Result:
The percentage change in the de Broglie wavelength when the kinetic energy is increased by 10 times is approximately **68.4%**.