If the kinetic energy of the particle is increased to `16` times its previous value , 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, we need to find the percentage change in the de Broglie wavelength of a particle when its kinetic energy is increased to 16 times its previous value.

### Step-by-Step Solution:

1. **Understand the de Broglie wavelength formula**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant and \( p \) is the momentum of the particle.

2. **Relate momentum to kinetic energy**:
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( E \):
\[ E = \frac{p^2}{2m} \implies p = \sqrt{2mE} \]
Substituting this into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

3. **Determine the new kinetic energy**:
If the kinetic energy is increased to 16 times its previous value, we can denote the original kinetic energy as \( E \) and the new kinetic energy as:
\[ E' = 16E \]

4. **Calculate the new de Broglie wavelength**:
Substitute \( E' \) into the de Broglie wavelength formula:
\[ \lambda' = \frac{h}{\sqrt{2mE'}} = \frac{h}{\sqrt{2m(16E)}} = \frac{h}{\sqrt{32mE}} = \frac{h}{4\sqrt{2mE}} = \frac{\lambda}{4} \]
Thus, the new wavelength \( \lambda' \) is \( \frac{\lambda}{4} \).

5. **Calculate the change in wavelength**:
The change in wavelength is:
\[ \Delta \lambda = \lambda - \lambda' = \lambda - \frac{\lambda}{4} = \frac{3\lambda}{4} \]

6. **Calculate the percentage change in wavelength**:
The percentage change in wavelength is given by:
\[ \text{Percentage change} = \frac{\Delta \lambda}{\lambda} \times 100\% = \frac{\frac{3\lambda}{4}}{\lambda} \times 100\% = \frac{3}{4} \times 100\% = 75\% \]

### Final Answer:
The percentage change in the de Broglie wavelength of the particle is **75%**.