A proton is accelerated through 225 V . Its de Broglie wavelength is

Options

A. 0.001 nm
B. 0.2 nm
C. 0.3 nm
D. 0.4 nm

Answer

To find the de Broglie wavelength of a proton accelerated through a potential difference of 225 V, we can follow these steps:

### Step 1: Understand the formula for de Broglie wavelength
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 \( p \) can be expressed in terms of kinetic energy (K.E.) as follows:
\[ p = \sqrt{2m \cdot \text{K.E.}} \]
where \( m \) is the mass of the proton.

### Step 3: Express kinetic energy in terms of charge and voltage
The kinetic energy gained by the proton when accelerated through a potential difference \( V \) is given by:
\[ \text{K.E.} = q \cdot V \]
where \( q \) is the charge of the proton.

### Step 4: Substitute kinetic energy in the momentum formula
Substituting the expression for kinetic energy into the momentum formula gives:
\[ p = \sqrt{2m \cdot (q \cdot V)} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot (q \cdot V)}} \]

### Step 6: Insert known values
We will now insert the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Voltage \( V = 225 \, \text{V} \)

### Step 7: Calculate the de Broglie wavelength
Substituting these values into the equation:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot 225}} \]

Calculating the denominator:
1. Calculate \( 2m \cdot (q \cdot V) \):
\[ 2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot 225 \]
This results in a numerical value.

2. Take the square root of that value.

3. Finally, calculate \( \lambda \).

### Step 8: Final result
After performing the calculations, we find:
\[ \lambda \approx 0.194 \times 10^{-11} \, \text{m} = 0.00194 \, \text{nm} \]

### Conclusion
Thus, the de Broglie wavelength of the proton accelerated through 225 V is approximately \( 0.00194 \, \text{nm} \).

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