Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse also true. This is summed up in what we now call the Heisenberg uncertainty principle.
The equation si `deltax.delta (mv)ge(h)/(4pi)`
The uncertainty in the position or in the momentum of a marcroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electon is small enough for the uncertainty to be relatively large and significant.
What would be the minimum uncetaintty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncetainty in position is `(7)/(22)` nm?

Options

A. `6.25 Å`
B. `6 Å`
C. `0.625 Å`
D. `0.1325 Å`

Explanation

To solve the problem, we will use the Heisenberg uncertainty principle and the relationship between momentum and de Broglie wavelength. Here are the steps:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle states that:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).

### Step 2: Given Values
From the problem, we have:
- \(\Delta x = \frac{7}{22} \, \text{nm} = \frac{7 \times 10^{-9}}{22} \, \text{m} \approx 3.18 \times 10^{-10} \, \text{m}\)
- Potential difference \(V = 6 \, \text{V}\)

### Step 3: Calculate the Momentum of the Electron
The kinetic energy gained by the electron when accelerated through a potential difference \(V\) is given by:
\[ KE = eV \]
where \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)). Therefore, the kinetic energy is:
\[ KE = (1.6 \times 10^{-19} \, \text{C})(6 \, \text{V}) = 9.6 \times 10^{-19} \, \text{J} \]
The momentum \(p\) of the electron can be calculated using the relation:
\[ p = \sqrt{2m \cdot KE} \]
where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)). Thus,
\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (9.6 \times 10^{-19} \, \text{J})} \]

### Step 4: Calculate the Uncertainty in Momentum
Using the Heisenberg uncertainty principle:
\[ \Delta p = \frac{h}{4\pi \Delta x} \]
Substituting the values:
\[ \Delta p = \frac{6.626 \times 10^{-34} \, \text{Js}}{4\pi \cdot (3.18 \times 10^{-10} \, \text{m})} \]

### Step 5: Relate Momentum to Wavelength
The de Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{p} \]
The uncertainty in wavelength \(\Delta \lambda\) can be related to \(\Delta p\):
\[ \Delta p = -\frac{h}{\lambda^2} \Delta \lambda \]
Thus,
\[ \Delta \lambda = -\frac{\lambda^2 \Delta p}{h} \]

### Step 6: Substitute and Solve for \(\Delta \lambda\)
Now substituting \(\Delta p\) into the equation for \(\Delta \lambda\):
\[ \Delta \lambda = -\frac{\lambda^2}{h} \cdot \frac{h}{4\pi \Delta x} \]
This simplifies to:
\[ \Delta \lambda = -\frac{\lambda^2}{4\pi \Delta x} \]

### Step 7: Calculate \(\Delta \lambda\)
We need to calculate \(\lambda\) first. The de Broglie wavelength for an electron accelerated through \(6 \, \text{V}\) can be approximated as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]
Substituting values, we can find \(\lambda\) and then use it to find \(\Delta \lambda\).

### Final Calculation
After performing the calculations, we find:
\[ \Delta \lambda \approx 0.625 \, \text{Å} \]

### Conclusion
Thus, the minimum uncertainty in the de Broglie wavelength of the moving electron is approximately \(0.625 \, \text{Å}\).

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