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

\(Δv = 2.4 \times 10-26 \ ms^{-1}\)
\(Δx = 10^{-7} m\)
By uncertainty principle,
\(∴ m ≥ \frac {h}{4\pi(Δx)(Δv)}\)

\(m≥ \frac {6.626 \times 10^{-34}}{4\times 3.14 \times (10^{-7})(2.4)\times 10^{-26}}\)

\(m≥ \frac {6.626 \times 10^{-1}}{4 \times 2.4\times3.14}\)

\(m≥ 0.02198\ kg\)
\(m≥ 21.98\ g\)

So, the mass of the particle \(≃ 22\ g\)