If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is

Options

A. (a)`25%`
B. (b)`75%`
C. (c)`60%`
D. (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 by 16 times. Let's go through the solution step by step.

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity.

### Step 2: Relate kinetic energy to velocity
The kinetic energy (\(E\)) of a particle is given by:
\[ E = \frac{1}{2}mv^2 \]
From this equation, we can express the velocity in terms of kinetic energy:
\[ v = \sqrt{\frac{2E}{m}} \]

### Step 3: Substitute velocity into the de Broglie wavelength formula
Substituting \(v\) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m\sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2mE}} \]

### Step 4: Find the initial wavelength
Let the initial kinetic energy be \(E\). The initial de Broglie wavelength (\(\lambda_1\)) is:
\[ \lambda_1 = \frac{h}{\sqrt{2mE}} \]

### Step 5: Determine the new kinetic energy
If the kinetic energy is increased by 16 times, the new kinetic energy (\(E'\)) is:
\[ E' = 16E \]

### Step 6: Find the new wavelength
The new de Broglie wavelength (\(\lambda_2\)) with the increased kinetic energy is:
\[ \lambda_2 = \frac{h}{\sqrt{2mE'}} = \frac{h}{\sqrt{2m(16E)}} = \frac{h}{\sqrt{32mE}} = \frac{h}{4\sqrt{2mE}} = \frac{\lambda_1}{4} \]

### Step 7: Calculate the percentage change in wavelength
The percentage change in wavelength is given by:
\[ \text{Percentage Change} = \frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100 \]
Substituting \(\lambda_2\):
\[ \text{Percentage Change} = \frac{\lambda_1 - \frac{\lambda_1}{4}}{\lambda_1} \times 100 = \frac{\lambda_1 - \frac{\lambda_1}{4}}{\lambda_1} \times 100 = \frac{3\lambda_1/4}{\lambda_1} \times 100 = \frac{3}{4} \times 100 = 75\% \]

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

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