What would be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncertainty in position is `(7)/(22)` nm?

Options

A. `6.25 Å`
B. `6 Å`
C. `0.625 Å`
D. `0.3125 Å`

Answer

To find the minimum uncertainty in the de-Broglie wavelength of a moving electron accelerated by a potential difference of 6 volts, given the uncertainty in position is \( \frac{7}{22} \) nm, we can follow these steps:

### Step 1: Calculate the de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) of an electron accelerated through a potential difference \( V \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of the potential difference as:
\[ p = \sqrt{2m_e eV} \]
where:
- \( m_e \) is the mass of the electron (\( 9.11 \times 10^{-31} \) kg),
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C),
- \( V \) is the potential difference (6 V).

### Step 2: Substitute values to find momentum
Substituting the values into the momentum equation:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31} \, \text{kg}) \times (1.6 \times 10^{-19} \, \text{C}) \times 6 \, \text{V}} \]

Calculating this gives:
\[ p \approx \sqrt{1.742 \times 10^{-48}} \approx 1.32 \times 10^{-24} \, \text{kg m/s} \]

### Step 3: Calculate the de-Broglie wavelength
Now, substituting \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) (Planck's constant) is \( 6.626 \times 10^{-34} \, \text{Js} \).

Calculating \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.32 \times 10^{-24}} \approx 5.01 \times 10^{-10} \, \text{m} \approx 5 \, \text{Å} \]

### Step 4: Apply Heisenberg's Uncertainty Principle
According to Heisenberg's uncertainty principle:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where \( \Delta x \) is the uncertainty in position and \( \Delta p \) is the uncertainty in momentum.

### Step 5: Calculate uncertainty in momentum
We know:
\[ \Delta p = \frac{h}{\lambda^2} \Delta \lambda \]
Substituting this into the uncertainty principle:
\[ \Delta x \cdot \frac{h}{\lambda^2} \Delta \lambda \geq \frac{h}{4\pi} \]
Cancelling \( h \) from both sides:
\[ \Delta x \cdot \Delta \lambda \geq \frac{\lambda^2}{4\pi} \]

### Step 6: Solve for \( \Delta \lambda \)
Rearranging gives:
\[ \Delta \lambda \geq \frac{\lambda^2}{4\pi \Delta x} \]

Substituting \( \lambda \approx 5 \times 10^{-10} \, \text{m} \) and \( \Delta x = \frac{7}{22} \, \text{nm} = \frac{7 \times 10^{-9}}{22} \, \text{m} \):
\[ \Delta \lambda \geq \frac{(5 \times 10^{-10})^2}{4\pi \left(\frac{7 \times 10^{-9}}{22}\right)} \]

Calculating this gives:
\[ \Delta \lambda \geq \frac{25 \times 10^{-20}}{4\pi \times \frac{7 \times 10^{-9}}{22}} \approx 0.625 \, \text{Å} \]

### Final Answer
Thus, the minimum uncertainty in the de-Broglie wavelength is:
\[ \Delta \lambda \geq 0.625 \, \text{Å} \]

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