Question

An electron is confi ned to a linear region with a length of the same order as the diameter of an atom (about 100 pm). Calculate the minimum uncertainties in its position and speed.

Answer 1

Step-by-step explanation:

The minimum uncertainty position is 100 nm.

Therefore, by Heisenberg Uncertainty principal

ΔxΔp≥h/2

Δp≥h/(2Δx)

we know that

h=
`1.0546*10^(-34)` Js

Δx=
`100*10^(-12) m`

therefore,

`\Delta p =(1.0546*10^(-34))/(2*12*10^(-12)) = 5.3*10^(-25) kgms^(-1)`

therefore,

`\Delta v= (\Delta p)/(m) =(5.3*10^(-25))/(9.11*10^(-31))= 5.8*10^(5) ms^(-1)`