Question

If the uncertainty in velocity and position of a minute particle in space are, \(2.4 × 10^{–26}\) \((m s^{–1)}\) and \(10^{–7} (m)\), respectively. The mass of the particle in g is _____ . (Nearest integer)
(Given : \(h = 6.626 × 10^{–34} Js\))

Answer 1

Correct Answer: 22

To solve the problem using the Heisenberg Uncertainty Principle, we apply the following relation: \(\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\), where \(\Delta x\) is the uncertainty in position, \(\Delta p\) is the uncertainty in momentum, and \(h\) is Planck's constant.

Given:

• \(\Delta v = 2.4 \times 10^{-26}\) m/s (uncertainty in velocity)
• \(\Delta x = 10^{-7}\) m (uncertainty in position)
• Planck’s constant \(h = 6.626 \times 10^{-34}\) Js

We know that momentum \(p\) is given by \(p = m \cdot v\), thus \(\Delta p = m \cdot \Delta v\).

Substitute the known values into the uncertainty principle:

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

\[10^{-7} \cdot m \cdot 2.4 \times 10^{-26} \geq \frac{6.626 \times 10^{-34}}{4\pi}\]

Solve for \(m\):

\[m \geq \frac{6.626 \times 10^{-34}}{4\pi \cdot 10^{-7} \cdot 2.4 \times 10^{-26}}\]

Calculate the right side:

\[m \geq \frac{6.626 \times 10^{-34}}{3.019 \times 10^{-32}}\]

\[m \geq 21.942 \approx 22 \text{ kg}\]

Convert the mass from kg to g (1 kg = 1000 g):

\[m \approx 22000 \text{ g}\]

The given range is 22, meaning that rounding to the nearest integer, the mass of the particle is 22 g.

The computed solution, 22, is within the specified range of 22, validating our solution.