Calculate the de-Broglie wavelength of an electron beam accelerated through a potential difference of 60 V.

Options

A. 1.58 A
B. 2.58 A
C. 3.58 A
D. 4.58. A .

Answer

To calculate the de-Broglie wavelength of an electron beam accelerated through a potential difference of 60 V, 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 electron.

### Step 2: Relate momentum to kinetic energy
The kinetic energy (\( KE \)) of the electron when accelerated through a potential difference \( V \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron. The momentum \( p \) can be expressed in terms of kinetic energy as:
\[ p = \sqrt{2m \cdot KE} \]
Substituting \( KE \) into this equation gives:
\[ p = \sqrt{2m \cdot eV} \]

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Now we can substitute the expression for momentum into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 4: Insert known values
Now, we need to insert the known values:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Charge of the electron \( e = 1.602 \times 10^{-19} \, \text{C} \)
- Potential difference \( V = 60 \, \text{V} \)

Substituting these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot (9.11 \times 10^{-31}) \cdot (1.602 \times 10^{-19}) \cdot 60}} \]

### Step 5: Calculate the denominator
First, calculate the denominator:
\[ 2 \cdot (9.11 \times 10^{-31}) \cdot (1.602 \times 10^{-19}) \cdot 60 = 1.094 \times 10^{-48} \]
Now, take the square root:
\[ \sqrt{1.094 \times 10^{-48}} \approx 1.046 \times 10^{-24} \]

### Step 6: Calculate the de-Broglie wavelength
Now substitute back into the equation for \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.046 \times 10^{-24}} \approx 6.34 \times 10^{-10} \, \text{m} \approx 0.634 \, \text{nm} \]

### Step 7: Convert to angstroms
To convert to angstroms (1 nm = 10 Å):
\[ \lambda \approx 6.34 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the electron beam accelerated through a potential difference of 60 V is approximately \( 6.34 \, \text{Å} \).

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