What is the percentage change in the de-Broglie wavelength of the particle if its kinetic energy is increased 15 times?

Explanation

To find the percentage change in the de-Broglie wavelength of a particle when its kinetic energy is increased 15 times, we can follow these steps:

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

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) can be expressed in terms of kinetic energy (\( K \)):
\[ K = \frac{p^2}{2m} \implies p = \sqrt{2mK} \]
Substituting this into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 3: Determine the initial wavelength
Let the initial kinetic energy be \( K \) and the initial wavelength be \( \lambda \):
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 4: Calculate the new kinetic energy
If the kinetic energy is increased 15 times, the new kinetic energy \( K' \) is:
\[ K' = 15K \]

### Step 5: Calculate the new wavelength
Now, substituting \( K' \) into the wavelength formula:
\[ \lambda' = \frac{h}{\sqrt{2mK'}} = \frac{h}{\sqrt{2m(15K)}} = \frac{h}{\sqrt{15} \cdot \sqrt{2mK}} = \frac{1}{\sqrt{15}} \cdot \frac{h}{\sqrt{2mK}} = \frac{\lambda}{\sqrt{15}} \]

### Step 6: Find the percentage change in wavelength
The percentage change in wavelength is given by:
\[ \text{Percentage Change} = \frac{\lambda - \lambda'}{\lambda} \times 100 \]
Substituting \( \lambda' \):
\[ \text{Percentage Change} = \frac{\lambda - \frac{\lambda}{\sqrt{15}}}{\lambda} \times 100 = \left(1 - \frac{1}{\sqrt{15}}\right) \times 100 \]

### Step 7: Simplify the expression
Calculating \( 1 - \frac{1}{\sqrt{15}} \):
\[ \sqrt{15} \approx 3.872 \implies \frac{1}{\sqrt{15}} \approx 0.258 \]
Thus,
\[ 1 - \frac{1}{\sqrt{15}} \approx 1 - 0.258 \approx 0.742 \]
Now, multiplying by 100 gives:
\[ \text{Percentage Change} \approx 0.742 \times 100 \approx 74.2\% \]

### Final Result
The percentage change in the de-Broglie wavelength when the kinetic energy is increased 15 times is approximately:
\[ \text{Percentage Change} \approx 74.2\% \]