Question

The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultaneous measurement of the momentum component px of this electron?

Answer 1

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$.