An electron is accelerated under a potential difference of 64 V, the de-Brogile wavelength associated with electron is `[e = -1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31)kg, h = 6.623 xx 10^(-34) Js]`

Options

A. `1. 53 Å`
B. `2. 53 Å`
C. `3. 35 Å`
D. `4. 54 Å`

Answer

To solve the problem of finding the de Broglie wavelength associated with an electron accelerated through a potential difference of 64 V, we will follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle 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 \( KE \):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
For an electron accelerated through a potential difference \( V \), the kinetic energy gained is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 3: Substitute kinetic energy into the momentum equation
Substituting \( KE \) into the momentum equation gives:
\[ p = \sqrt{2m \cdot eV} \]

### Step 4: Substitute momentum into the de Broglie wavelength formula
Now substituting \( p \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 5: Plug in the known values
Now we can substitute the known values:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( V = 64 \, \text{V} \)

Substituting these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 64}} \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ 2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 64 = 1.84768 \times 10^{-48} \]
Taking the square root:
\[ \sqrt{1.84768 \times 10^{-48}} \approx 1.360 \times 10^{-24} \]

### Step 7: Calculate the wavelength
Now substituting back into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.360 \times 10^{-24}} \approx 4.87 \times 10^{-10} \, \text{m} = 0.487 \, \text{nm} \]

### Step 8: Convert to Angstroms
Since \( 1 \, \text{nm} = 10 \, \text{Å} \):
\[ \lambda \approx 4.87 \, \text{Å} \]

### Final Result
The de Broglie wavelength associated with the electron is approximately \( 4.87 \, \text{Å} \).