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

Options

A. 0.1 nm
B. 0.2 nm
C. 0.3 nm
D. 0.4nm

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 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: Determine the momentum of the proton
When a proton is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
For a proton, the charge \( q \) is equal to the elementary charge \( e \) (approximately \( 1.6 \times 10^{-19} \) C). Therefore, the kinetic energy gained by the proton when accelerated through 225 V is:
\[ KE = e \cdot 225 \]

### Step 3: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the proton. We can equate the two expressions for kinetic energy:
\[ e \cdot 225 = \frac{p^2}{2m} \]

### Step 4: Solve for momentum
Rearranging the equation gives:
\[ p^2 = 2m \cdot e \cdot 225 \]
Taking the square root, we find:
\[ p = \sqrt{2m \cdot e \cdot 225} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot e \cdot 225}} \]

### Step 6: Insert known values
Using the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \) J·s
- Mass of the proton \( m = 1.67 \times 10^{-27} \) kg
- Charge of the proton \( e = 1.6 \times 10^{-19} \) C
- Potential difference \( V = 225 \) V

We can calculate:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot 225}} \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ 2 \cdot (1.67 \times 10^{-27}) \cdot (1.6 \times 10^{-19}) \cdot 225 \approx 1.15 \times 10^{-22} \]
Taking the square root:
\[ \sqrt{1.15 \times 10^{-22}} \approx 1.07 \times 10^{-11} \]

### Step 8: Calculate the wavelength
Now substituting back to find \( \lambda \):
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{1.07 \times 10^{-11}} \approx 6.19 \times 10^{-23} \text{ m} \]
Converting to nanometers:
\[ \lambda \approx 0.619 \text{ nm} \text{ or } 6.19 \text{ Å} \]

### Final Answer
The de Broglie wavelength of the proton is approximately \( 0.619 \) nm or \( 6.19 \) Å.

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