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

Answer

To solve the problem of determining 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 1: Understand the relationship between kinetic energy and de Broglie wavelength
The de Broglie wavelength (λ) of a particle 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 \(p\) of a particle is related to its kinetic energy (KE) by the equation:
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the particle. Rearranging this gives:
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Substitute kinetic energy into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Calculate the new de Broglie wavelength after increasing kinetic energy
If the kinetic energy is increased to 16 times its previous value, we can denote the initial kinetic energy as \(KE_0\) and the new kinetic energy as:
\[ KE_{new} = 16 \cdot KE_0 \]
Substituting this into the de Broglie wavelength formula gives:
\[ \lambda_{new} = \frac{h}{\sqrt{2m \cdot (16 \cdot KE_0)}} = \frac{h}{\sqrt{16 \cdot 2m \cdot KE_0}} = \frac{h}{4\sqrt{2m \cdot KE_0}} = \frac{\lambda_0}{4} \]
where \(\lambda_0\) is the original de Broglie wavelength.

### Step 5: Calculate the percentage change in wavelength
The percentage change in wavelength can be calculated using the formula:
\[ \text{Percentage Change} = \frac{\text{New Wavelength} - \text{Old Wavelength}}{\text{Old Wavelength}} \times 100 \]
Substituting the values we have:
\[ \text{Percentage Change} = \frac{\frac{\lambda_0}{4} - \lambda_0}{\lambda_0} \times 100 = \frac{-\frac{3\lambda_0}{4}}{\lambda_0} \times 100 = -75\% \]

### Final Answer
The percentage change in the de Broglie wavelength of the particle is \(-75\%\).

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