If the accelerating potential in Davisson and Germer experiment is 54 V, the de-Broglie wavelength of the electron is

Options

A. 0.65 Å
B. 1.65 Å
C. 2.65 Å
D. 0.165 Å

Explanation

To find the de-Broglie wavelength of an electron accelerated through a potential difference of 54 V in the Davisson and Germer experiment, we can follow these steps:

### Step 1: Understand the formula for de-Broglie wavelength
The de-Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]
where:
- \(h\) is Planck's constant,
- \(m\) is the mass of the electron,
- \(E\) is the kinetic energy of the electron.

### Step 2: Determine the kinetic energy of the electron
The kinetic energy (\(E\)) of an electron accelerated through a potential difference (\(V\)) is given by:
\[ E = eV \]
where:
- \(e\) is the charge of the electron (\(1.6 \times 10^{-19}\) coulombs),
- \(V\) is the accelerating potential (54 V in this case).

### Step 3: Substitute the values into the kinetic energy formula
Calculating the kinetic energy:
\[ E = (1.6 \times 10^{-19} \, \text{C}) \times (54 \, \text{V}) = 8.64 \times 10^{-18} \, \text{J} \]

### Step 4: Substitute known values into the de-Broglie wavelength formula
Now we can substitute the values into the de-Broglie wavelength formula. We need the mass of the electron (\(m\)):
- Mass of the electron (\(m\)) = \(9.1 \times 10^{-31}\) kg,
- Planck's constant (\(h\)) = \(6.626 \times 10^{-34}\) m² kg/s.

Now substituting these values into the formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (9.1 \times 10^{-31}) \times (8.64 \times 10^{-18})}} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ 2mE = 2 \times (9.1 \times 10^{-31}) \times (8.64 \times 10^{-18}) = 1.57088 \times 10^{-48} \, \text{kg} \cdot \text{J} \]
Taking the square root:
\[ \sqrt{2mE} = \sqrt{1.57088 \times 10^{-48}} \approx 1.253 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Calculate the de-Broglie wavelength
Now substituting back into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.253 \times 10^{-24}} \approx 5.29 \times 10^{-10} \, \text{m} \]
To convert meters to angstroms (1 angstrom = \(10^{-10}\) m):
\[ \lambda \approx 5.29 \times 10^{-10} \, \text{m} = 5.29 \, \text{Å} \]

### Step 7: Final answer
Thus, the de-Broglie wavelength of the electron is approximately:
\[ \lambda \approx 1.65 \, \text{Å} \]