Uncertainty in position of an electron ($m=9.1\times {10}^{-28}g$) 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 $h=6.626\times {10}^{-27}ergs$)

The correct option is C $1.93cm$
According to Heisenbergs uncertainity principle,

$\Delta x\cdot m\Delta v=\frac{h}{4\Pi }$

$so,\Delta x\cdot \Delta v=\frac{h}{4\Pi m}$

Hence, product of uncertainity in position and velocity=$\frac{h}{4\pi m}$

$Here,m=9.1\times {10}^{-28}g,h=6.626\times {10}^{-27}Js$

now, product of uncertainity in position and velocity$=6.626\times {10}^{-27}/(4\times 3.14\times {9.1\times 10}^{-28}\times 0.3)$

$=(6.626\times {10}^{-27})/(4\times 3.14\times 9.1\times {10}^{-27}\times 0.3)$

$=1.93cm$