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 of finding the percentage change in the de Broglie wavelength of a particle when its kinetic energy is increased to 16 times its previous value, 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.

2. **Relate Momentum to Kinetic Energy**:
The momentum \( p \) can be expressed in terms of mass \( m \) and velocity \( v \):
\[ p = mv \]
The kinetic energy (KE) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]

3. **Set Up the Initial and Final Kinetic Energy**:
Let the initial kinetic energy be \( KE = \frac{1}{2} mv^2 \).
If the kinetic energy is increased to 16 times its previous value, the new kinetic energy \( KE' \) is:
\[ KE' = 16 \times KE = 16 \times \frac{1}{2} mv^2 = 8mv^2 \]

4. **Find the New Velocity**:
The new kinetic energy can also be expressed in terms of the new velocity \( v' \):
\[ KE' = \frac{1}{2} mv'^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ 8mv^2 = \frac{1}{2} mv'^2 \]
Simplifying this equation:
\[ 16v^2 = v'^2 \implies v' = 4v \]

5. **Calculate the New Wavelength**:
Now, substituting \( v' \) back into the de Broglie wavelength formula:
\[ \lambda' = \frac{h}{p'} = \frac{h}{mv'} = \frac{h}{m(4v)} = \frac{1}{4} \frac{h}{mv} = \frac{1}{4} \lambda \]

6. **Determine the Change in Wavelength**:
The change in wavelength \( \Delta \lambda \) is given by:
\[ \Delta \lambda = \lambda' - \lambda = \frac{1}{4} \lambda - \lambda = -\frac{3}{4} \lambda \]

7. **Calculate the Percentage Change**:
The percentage change in wavelength is given by:
\[ \text{Percentage Change} = \frac{\Delta \lambda}{\lambda} \times 100 = \frac{-\frac{3}{4} \lambda}{\lambda} \times 100 = -75\% \]
This indicates a decrease in wavelength.

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