Question

Product of uncertainty in position (\(\Delta x\)) and uncertainty in velocity (\(\Delta v\)) has unit:

• A. \(m\cdot s\)
• B. \(m^2\cdot s^{-1}\)
• C. \(m\cdot s^{-1}\)
• D. \(m^2\cdot s^2\)

Hint:

The product of uncertainty in position and uncertainty in velocity has units of momentum, which are \(m \cdot s^{-1}\).

Answer 1

The Correct Option is C

The Heisenberg uncertainty principle relates the uncertainty in position (\(\Delta x\)) and the uncertainty in momentum (\(\Delta p\)), which can be expressed as: \[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \] Where: - \(\hbar\) is the reduced Planck constant. Momentum is defined as \(p = m \cdot v\), where \(m\) is mass and \(v\) is velocity. Therefore, the uncertainty in momentum is: \[ \Delta p = m \cdot \Delta v \] Substituting into the uncertainty principle: \[ \Delta x \cdot m \cdot \Delta v \geq \frac{\hbar}{2} \] This shows that the product of uncertainty in position and uncertainty in velocity has units of: \[ \Delta x \cdot \Delta v \sim m \cdot s^{-1} \] Thus, the correct unit for the product of \(\Delta x\) and \(\Delta v\) is \(m \cdot s^{-1}\).