Question

Calculate the uncertainty in the position of a particle when the uncertainty in momentum is
(a) $1\times {10}^{-3}$ $g$ $cm$ ${sec}^{-1}$ (b) zero.

Answer 1

(a) Given,
$\Delta P=1\times {10}^{-3}$ $g$ $cm$ ${sec}^{-1}$
$h=6.62\times {10}^{-27}erg-sec$
$\pi =3.142$
According to uncertainty principle,
$\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$
So, $\Delta x\ge \frac{h}{4\pi }\cdot \frac{1}{\Delta p}\ge \frac{6.62\times {10}^{-27}}{4\times 3.142}\times \frac{1}{{10}^{-3}}$
$=0.527\times {10}^{-24}cm$
(b) When the value of $\Delta p=0$, the value of $\Delta x$ will be infinity.