The de-Broglie wavelength associated with proton changes by `0.25%` if its momentum is changed by `p_(0)`. The initial momentum was

Options

A. `401 p_(0) `
B. `(P_(0))/(4)`
C. `5 p_(0)`
D. `(p_(0))/(5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between momentum and de-Broglie wavelength
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: Set up the initial and final conditions
Let the initial momentum of the proton be \( p \). The initial wavelength is:
\[ \lambda = \frac{h}{p} \]
When the momentum changes by \( p_0 \), the new momentum becomes \( p - p_0 \). The new wavelength (\( \lambda' \)) is:
\[ \lambda' = \frac{h}{p - p_0} \]

### Step 3: Express the change in wavelength
According to the problem, the wavelength changes by \( 0.25\% \). This means:
\[ \lambda' = \lambda + 0.0025 \lambda = \lambda (1 + 0.0025) \]
Thus, we can write:
\[ \frac{h}{p - p_0} = \frac{h}{p} (1 + 0.0025) \]

### Step 4: Simplify the equation
Cancelling \( h \) from both sides gives:
\[ \frac{1}{p - p_0} = \frac{1 + 0.0025}{p} \]
Cross-multiplying yields:
\[ p = (p - p_0)(1 + 0.0025) \]

### Step 5: Expand and rearrange
Expanding the right-hand side:
\[ p = p - p_0 + 0.0025p - 0.0025p_0 \]
Rearranging gives:
\[ p_0 = 0.0025p - 0.0025p_0 \]
Combining like terms:
\[ p_0 + 0.0025p_0 = 0.0025p \]
Factoring out \( p_0 \):
\[ p_0(1 + 0.0025) = 0.0025p \]

### Step 6: Solve for \( p \)
Now, we can isolate \( p \):
\[ p = \frac{p_0(1 + 0.0025)}{0.0025} \]
Calculating \( 1 + 0.0025 = 1.0025 \):
\[ p = \frac{1.0025}{0.0025} p_0 \]
Calculating \( \frac{1.0025}{0.0025} = 401 \):
\[ p = 401 p_0 \]

### Final Answer
The initial momentum \( p \) is:
\[ \boxed{401 p_0} \]