If the de-Broglie wavelength of a proton is `10^(-13)` m, the electric potentia through which it must have been accelerated is

Options

A. `4.07xx10^4 V`
B. `8.15xx10^4 V`
C. `8.15xx10^3 V`
D. `4.07xx10^5 V`

Explanation

To find the electric potential through which a proton must have been accelerated to achieve a de-Broglie wavelength of \(10^{-13}\) m, 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 electric potential
The momentum \(p\) of a proton accelerated through a potential \(V\) can be expressed as:
\[ p = \sqrt{2m_e \cdot e \cdot V} \]
where:
- \(m_e\) is the mass of the proton,
- \(e\) is the charge of the proton (which is equal to the charge of an electron, \(1.6 \times 10^{-19}\) C),
- \(V\) is the electric potential.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2m_e \cdot e \cdot V}} \]

### Step 4: Rearranging the equation to solve for \(V\)
Squaring both sides gives:
\[ \lambda^2 = \frac{h^2}{2m_e \cdot e \cdot V} \]
Rearranging this to solve for \(V\) gives:
\[ V = \frac{h^2}{2m_e \cdot e \cdot \lambda^2} \]

### Step 5: Substitute known values
Now we can substitute the known values:
- \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- \(m_e = 1.67 \times 10^{-27} \, \text{kg}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(\lambda = 10^{-13} \, \text{m}\)

Substituting these values into the equation:
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-13})^2} \]

### Step 6: Calculate \(V\)
Calculating the numerator:
\[ (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2 \text{s}^2 \]

Calculating the denominator:
\[ 2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-26}) = 5.34 \times 10^{-46} \, \text{kg C m}^2 \]

Now substituting these into the equation for \(V\):
\[ V = \frac{4.39 \times 10^{-67}}{5.34 \times 10^{-46}} \approx 8.21 \times 10^{4} \, \text{V} \]

### Final Answer
The electric potential through which the proton must have been accelerated is approximately:
\[ V \approx 8.21 \times 10^{4} \, \text{V} \]