Answer
$$\therefore \Delta x = 8\times 10^{-13}\text{ m}$$
Work Step by Step
$v = 4\times 10^6 \text{ m/s }\pm 1\text{ percent}$
$v = 4\times 10^6 \text{ m/s }\pm 4\times 10^4\text{ m/s}$
$\therefore \Delta v = 4\times 10^4\text{ m/s}$
$\therefore \Delta p = m\Delta v = 66.9\times 10^{-24}\text{ kg.m/s}$
From the Heisenberg uncertainty principle,
$$\Delta x\cdot\Delta p\ge \frac{h}{4\pi}$$
for the minimum uncertainty in position, we will use the equality,
$$\Delta x\cdot\Delta p= \frac{h}{4\pi}$$
$$\Delta x= \frac{h}{4\pi\Delta p}$$
$$\therefore \Delta x = 7.88\times 10^{-13}\text{ m}$$
$$\therefore \Delta x \approx 8\times 10^{-13}\text{ m}$$
