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: Understanding the Concept
Heisenberg's uncertainty principle says the position and momentum of a particle cannot both be known exactly. The product of their uncertainties has a minimum value.

Step 2: Key Formula or Approach
\[ \Delta x\cdot m\,\Delta v \ge \frac{h}{4\pi} \quad\Rightarrow\quad \Delta v = \frac{h}{4\pi m\,\Delta x} \]

Step 3: Detailed Explanation
Convert the position uncertainty: \(\Delta x = 50\ \text{pm} = 5\times10^{-11}\) m.
Denominator: \(4\pi m\Delta x = 4(3.142)(9.1\times10^{-31})(5\times10^{-11}) = 5.718\times10^{-40}\).
\[ \Delta v = \frac{6.63\times10^{-34}}{5.718\times10^{-40}} = 1.16\times10^{6}\ \text{m s}^{-1} \]
The other options do not follow from this calculation.

Final Answer:
The minimum uncertainty in the electron's velocity is \(1.16\times10^6\) m/s, option (B). \[ \boxed{1.16\times10^{6}\ \text{m s}^{-1}\ \text{(B)}} \]