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

$Δv = 2.4 \times 10-26 \ ms^{-1}$
$Δx = 10^{-7} m$
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 $≃ 22\ g$