Question

Uncertainty in position of an electron (mass of an electron is $=9.1 \times 10^{-28} \,g$ ) moving with a velocity of $3 \times 10^{4}\, cm / s$ accurate upto $0.001 \%$ will be (use $\frac{h}{4 \pi}$ in uncertainty expression where $h=6.626 \times 10^{-27}$ erg s)

• A. 1.93 cm
• B. 3.84 cm
• C. 5.76 cm
• D. 7.68 cm

Answer 1

Answer: (A)

According to Heisenberg's uncertainty principle
$\Delta x \times \Delta v=\frac{h}{4\pi m}$
here, $\Delta x$ = uncertainty in position
$ \Delta v$ = uncertainty in velocity
$h = $ Planck's constant $(6.626 \times 10^{-27} Js)$
$m =$ mass of electron $(9.1 \times 10^{-28} kg)$
Here, $\Delta v=0.001 \% $ of $3\times10^4$
$=\frac{0.001}{100}\times3\times 10^4=0.3 \,cm /s$
$\therefore \Delta x=\frac{h}{4\pi m \Delta v}$
$=\frac{6.626 \times10^{-27}}{4\times3.14 \times 9.1\times10^{-28} \times 0.3}=1.93 \,cm$