If kinetic energy of a proton is increased nine times the wavelength of the de - Broglie wave associated with it would become :-

Options

A. 3 times
B. 9 times
C. `1/3` times
D. `1/9` times

Answer

To solve the problem, we need to understand the relationship between the de Broglie wavelength and the kinetic energy of a proton. Here’s a step-by-step solution:

### Step 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.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) of a particle is related to its mass \( m \) and velocity \( v \) by:
\[ p = mv \]
The kinetic energy (KE) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]

### Step 3: Express momentum in terms of kinetic energy
We can express \( v \) in terms of kinetic energy:
From \( KE = \frac{1}{2} mv^2 \), we can rearrange it to find \( v \):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
Substituting this into the momentum equation gives:
\[ p = m \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{2m \cdot KE} \]

### Step 4: Substitute momentum back into the de Broglie wavelength formula
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 5: Analyze the effect of increasing kinetic energy
If the kinetic energy of the proton is increased nine times, we denote the initial kinetic energy as \( KE_1 \) and the new kinetic energy as \( KE_2 \):
\[ KE_2 = 9 \cdot KE_1 \]

### Step 6: Find the new wavelength
Substituting \( KE_2 \) into the wavelength formula:
\[ \lambda_2 = \frac{h}{\sqrt{2m \cdot KE_2}} = \frac{h}{\sqrt{2m \cdot (9 \cdot KE_1)}} = \frac{h}{\sqrt{9 \cdot 2m \cdot KE_1}} = \frac{h}{3\sqrt{2m \cdot KE_1}} = \frac{1}{3} \cdot \lambda_1 \]

### Step 7: Conclusion
Thus, when the kinetic energy of the proton is increased nine times, the new wavelength \( \lambda_2 \) becomes:
\[ \lambda_2 = \frac{1}{3} \cdot \lambda_1 \]

### Final Answer
The wavelength of the de Broglie wave associated with the proton becomes one-third of the original wavelength.

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