Answer:
Approximately at least [tex]5.73\times 10^{-10}\; \rm m[/tex].
Explanation:
Look up the following values:
The uncertainty in the electron's velocity is [tex]\Delta v = 1.01 \times 10^{5}\; \rm m \cdot s^{-1}[/tex]. Momentum is equal to mass times velocity. Hence, the uncertainty in the electron's momentum would be
[tex]\begin{aligned}\Delta p &= m \cdot \Delta v \\ &\approx 9.109 \times 10^{-31} \;\rm kg \times 1.01 \times 10^5\; \rm m \cdot s^{-1} \\ &\approx 9.20 \times 10^{-26}\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].
By Heisenberg's Uncertainty Principle, the product of uncertainty in an object's momentum [tex]\Delta p[/tex] and uncertainty in its position [tex]\Delta x[/tex] is at least [tex]\displaystyle \frac{h}{4\, \pi}[/tex].
That is:
[tex]\displaystyle \Delta p \cdot \Delta x \ge \frac{h}{4\, \pi}[/tex].
Rearrange to obtain:
[tex]\begin{aligned} \Delta x \ge \frac{h}{4\, \pi\, \Delta p} &\approx\frac{6.626\times 10^{-34}\; \rm kg \cdot m^2 \cdot s^{-1}}{(4\,\pi)\times 9.20\times 10^{-26}\; \rm kg \cdot m \cdot s^{-1}} \approx 5.73 \times 10^{-10}\; \rm m\end{aligned}[/tex].
In other words, the uncertainty in this electron's position is at least [tex]5.73\times 10^{-10}\; \rm m[/tex].
