For a proton accelerated through V volts , de Broglie wavelength is given as ` lambda = `

Options

A. `(12.27)/(sqrt(V)) Å`
B. `(0.101)/(sqrt(V)) Å `
C. ` (0.286 )/(sqrt(V)) Å `
D. `(12400)/(V) Å `

Answer

To find the de Broglie wavelength of a proton accelerated through a potential difference of V volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding 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.

2. **Momentum of the Proton**:
The momentum \( P \) of the proton can be expressed as:
\[ P = mv \]
where \( m \) is the mass of the proton and \( v \) is its velocity.

3. **Kinetic Energy from the Potential Difference**:
When a proton is accelerated through a potential difference \( V \), the work done on the proton is equal to the kinetic energy gained:
\[ KE = qV \]
where \( q \) is the charge of the proton. For a proton, \( q = 1.6 \times 10^{-19} \, \text{C} \).

4. **Relating Kinetic Energy to Momentum**:
The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{P^2}{2m} \]
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{P^2}{2m} \]

5. **Solving for Momentum**:
Rearranging the above equation for momentum \( P \):
\[ P^2 = 2mqV \]
Taking the square root:
\[ P = \sqrt{2mqV} \]

6. **Substituting Momentum into the de Broglie Wavelength Formula**:
Now substituting \( P \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

7. **Inserting Values**:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \, \text{C} \)

Thus, we can write:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (1.67 \times 10^{-27}) \times (1.6 \times 10^{-19}) \times V}} \]

8. **Final Expression**:
After calculating the constants, we find:
\[ \lambda = \frac{2.86 \times 10^{-11}}{\sqrt{V}} \, \text{m} \]
or, converting to angstroms:
\[ \lambda = \frac{0.286}{\sqrt{V}} \, \text{Å} \]

### Final Answer:
The de Broglie wavelength of a proton accelerated through \( V \) volts is:
\[ \lambda = \frac{0.286}{\sqrt{V}} \, \text{Å} \]