Question

Uncertainty in position of an electron (mass of an electron is =$9.1\times10^{-28}g)$ 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
• C. 5.76 cm
• D. 7.68

Answer 1

Answer: (A)

1.93 cm

Answer 2

According to Heisenberg's uncertainty principle
\hspace15mm \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, $\Delta v=0.001$% $of 3\times10^4$
$\, \, \, \, \, \, \, \, \, \, \, \, =\frac{0.001}{100}\times3\times 10^4=0.3 cm /s$
$\therefore\, \, \, \, \Delta x=\frac{h}{4\pi m \Delta v}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{6.626 \times10^{-27}}{4\times3.14 \times 9.1\times10^{-28} \times 0.3}=1.93 cm$