The uncertainly in position for an electron is `(lamda)/(4 pi)` where `lamda` is the de Broglie wavelength. The uncertainly in velocity will be -

Options

A. `(V)/(2)`
B. `V`
C. 3 V
D. `(V)/(4)`

Explanation

To find the uncertainty in velocity (Δv) given the uncertainty in position (Δx) for an electron, we can use the Heisenberg Uncertainty Principle, which states:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

Where:
- Δx = uncertainty in position
- Δp = uncertainty in momentum
- h = Planck's constant

Since momentum (p) is defined as the product of mass (m) and velocity (v), we can express the uncertainty in momentum as:

\[ \Delta p = m \cdot \Delta v \]

Substituting this into the uncertainty principle gives us:

\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

Now, we know from the problem statement that:

\[ \Delta x = \frac{\lambda}{4\pi} \]

where \(\lambda\) is the de Broglie wavelength, given by:

\[ \lambda = \frac{h}{m v} \]

Substituting this expression for Δx into the uncertainty principle:

\[ \frac{\lambda}{4\pi} \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

Now substituting \(\lambda\):

\[ \frac{\frac{h}{mv}}{4\pi} \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

Simplifying this:

\[ \frac{h \cdot \Delta v}{4\pi} \geq \frac{h}{4\pi} \]

Now, we can cancel \( \frac{h}{4\pi} \) from both sides (assuming \(h\) is not zero):

\[ \Delta v \geq 1 \]

Thus, the uncertainty in velocity (Δv) is:

\[ \Delta v \geq \frac{1}{m} \cdot \frac{h}{\lambda} \]

### Final Answer:
The uncertainty in velocity will be greater than or equal to \(\frac{h}{4\pi m}\).

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