Question

An electron has a speed of $50\, m / s$ with uncertainty of $0.02 \%$. The uncertainty in locating its position is

• A. $1.1 \times 10^{-1} \, m$
• B. $3.2 \times 10^{-7} \, m$
• C. $1.2 \times 10^{-5} \, m$
• D. $5.8 \times 10^{-3} \, m$

Answer 1

Answer: (D)

$\Delta v =\frac{0.02}{100} \times 50=0.01$
According of Heisenberg's uncertainty principle,
$ \Delta x \times \Delta v =\frac{ h }{4 \pi m } $
$ \Delta x =\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.01} $
$=5.77 \times 10^{-3} \,m $
$\approx 5.8 \times 10^{-3} \,m$