Question

Uncertainty in the position of an electron (mass = $9.1\times {10}^{-31}kg$) moving with a velocity of 100 m/s accurate upto 0.0001% will be:

• A. $5.79$ $m$
• B. $0.579$ $m$
• C. $5.79\times {10}^{-2}$ $m$
• D. $5.79\times {10}^{-3}$ $m$

Answer 1

The correct option is B $0.579$ $m$

We know that from Heisenberg's Uncertainity principle,
$\Delta x$ $\times$ $\Delta p$ $\ge$ $\frac{h}{4\pi }$
where $\Delta x$ is uncertainity in position
$\Delta p$ is uncertainity in momentum
h is Planck's constant
$Here,$ $p$ $=$ $mv$
where $p$ = momentumof particle
$m$ = mass of particle
$v$ = velocity of particle.
$\Delta p$ $=$ $\Delta \left(mv\right)$
$\Delta p$ $=$ $m\left(\Delta v\right)$ ; since m is constant.
$\Delta x_{min}$ $=$ $\frac{h}{4\pi m\times \Delta v}$

$=$ $\frac{6.626\times {10}^{-34}}{4\pi \times (9.1\times {10}^{-31})\times (100\times \frac{0.0001}{100})}$
$=$ $0.579$ $m$