Question

An electron is located on a pinpoint having a diameter of 3.52 µm. What is the minimum uncertainty in the speed of the electron?

Answer 1

Explanation:

It is given that,

An electron is located on a pinpoint having a diameter of 3.52 µm, $\Delta x=3.52\times 10^{-6}\ m$

We need to find the minimum uncertainty in the speed of the electron. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x \geq \dfrac{h}{4\pi}$

$\Delta p \geq \dfrac{h}{4\pi \Delta x}$

Since, p = m v

So, $mv \geq \dfrac{h}{4\pi \Delta x}$

$\Delta v \geq \dfrac{h}{4\pi \Delta x m}$

$\Delta v \geq \dfrac{6.67\times 10^{-34}}{4\pi 3.52\times 10^{-6}\times 9.1\times 10^{-31}}$

$\Delta v\geq 16.57\ m/s$

So, the minimum uncertainty in the speed of the electron is greater than 16.57 m/s. Hence, this is the required solution.