Explanation:
It is given that,
An electron is located on a pinpoint having a diameter of 3.52 µm, [tex]\Delta x=3.52\times 10^{-6}\ m[/tex]
We need to find the minimum uncertainty in the speed of the electron. It can be calculated using Heisenberg uncertainty principal as :
[tex]\Delta p.\Delta x \geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta p \geq \dfrac{h}{4\pi \Delta x}[/tex]
Since, p = m v
So, [tex]mv \geq \dfrac{h}{4\pi \Delta x}[/tex]
[tex]\Delta v \geq \dfrac{h}{4\pi \Delta x m}[/tex]
[tex]\Delta v \geq \dfrac{6.67\times 10^{-34}}{4\pi 3.52\times 10^{-6}\times 9.1\times 10^{-31}}[/tex]
[tex]\Delta v\geq 16.57\ m/s[/tex]
So, the minimum uncertainty in the speed of the electron is greater than 16.57 m/s. Hence, this is the required solution.
