Question

What is uncertainty in velocity of an electron if uncertainty in measurement of position is 50 pm ? \((m_e = 9.1\times 10^{-31}\text{kg}, h = 6.63\times 10^{-34}\text{Js}, π = 3.142)\)

• A. \(0.98\times 10^6 \text{ms}^{-1}\)
• B. \(1.16\times 10^6 \text{ms}^{-1}\)
• C. \(2.61\times 10^6 \text{ms}^{-1}\)
• D. \(3.77\times 10^6 \text{ms}^{-1}\)

Hint:

Use Heisenberg's relation \(\Delta v = h/(4\pi m\Delta x)\) with \(\Delta x = 5\times10^{-11}\) m.

Answer 1

The Correct Option is B

Step 1: Write the momentum form
Minimum uncertainty: $m\,\Delta v\,\Delta x = \dfrac{h}{4\pi}$. Evaluate the top and the bottom separately.

Step 2: Evaluate
Top: $h = 6.63\times10^{-34}$. Bottom: $4\pi = 12.568$ and $m\Delta x = 9.1\times10^{-31}\times5\times10^{-11} = 4.55\times10^{-41}$, so bottom $= 5.718\times10^{-40}$.
\[ \Delta v = \frac{6.63\times10^{-34}}{5.718\times10^{-40}} \approx 1.16\times10^{6}\ \text{m/s} \]
This matches option (B).

Final Answer:
The velocity uncertainty comes out as $1.16\times10^6$ m/s, option (B). \[ \boxed{1.16\times10^{6}\ \text{m/s}} \]