An electron has a speed of 40m/s, accurate up to 99.99%. what is the uncertainty in locating its position(in m)?

The correct option is A 0.0144m
According to Uncertainity principle

$\Delta$x. m$\Delta$v = $\frac{h}{4\pi }$

$\Delta$v = 40 m/s (Accurate upto $99.99\%$)

= $\frac{100-99.99}{100}$ x 40

= $0.004$ m/s

$m=9.1\times {10}^{-31}$ kg

$\Delta x$ = $\frac{h}{4\pi .m\Delta v}$

$\Delta x$ = $\frac{6.626\times {10}^{-34}}{4\times 3.14\times 9.1\times {10}^{-31}\times 0.004}$

$\Delta x$ = 0.0144 m