De-Broglie wavelengths of the particle increases by 75% then kinetic energy of particle becomes.

Options

A. `(16)/(49)` times
B. `(9)/(16)` times
C. `(16)/(25)`times
D. `(4)/(9)` times.

Explanation

To solve the problem, we need to analyze the relationship between the de Broglie wavelength (\(\lambda\)) and the kinetic energy (\(K\)) of a particle.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The de Broglie wavelength is related to the kinetic energy of a particle through the formula:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
where \(h\) is the Planck constant, \(m\) is the mass of the particle, and \(K\) is the kinetic energy.

2. **Expressing the Relationship**:
From the above formula, we can derive that:
\[ \lambda \propto \frac{1}{\sqrt{K}} \]
This implies that if the wavelength increases, the kinetic energy must decrease.

3. **Initial and Final Conditions**:
Let the initial de Broglie wavelength be \(\lambda_1\) and the initial kinetic energy be \(K_1\). According to the problem, the de Broglie wavelength increases by 75%, so the final wavelength \(\lambda_2\) can be expressed as:
\[ \lambda_2 = 1.75 \lambda_1 \]

4. **Setting Up the Ratio**:
Using the relationship derived earlier, we can set up the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} \]
Substituting \(\lambda_2 = 1.75 \lambda_1\) into the equation gives:
\[ \frac{\lambda_1}{1.75 \lambda_1} = \sqrt{\frac{K_2}{K_1}} \]
This simplifies to:
\[ \frac{1}{1.75} = \sqrt{\frac{K_2}{K_1}} \]

5. **Squaring Both Sides**:
Squaring both sides results in:
\[ \left(\frac{1}{1.75}\right)^2 = \frac{K_2}{K_1} \]
Calculating \(\left(\frac{1}{1.75}\right)^2\) gives:
\[ \frac{1}{3.0625} = \frac{K_2}{K_1} \]

6. **Simplifying the Ratio**:
To express this as a fraction, we can write:
\[ K_2 = K_1 \cdot \frac{1}{3.0625} = K_1 \cdot \frac{16}{49} \]

7. **Final Result**:
Thus, the kinetic energy of the particle after the increase in the de Broglie wavelength becomes:
\[ K_2 = \frac{16}{49} K_1 \]

### Conclusion:
The final kinetic energy of the particle after the increase in the de Broglie wavelength by 75% is \(\frac{16}{49}\) times the initial kinetic energy.