Question

An electron is confined 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

To solve this problem we will apply the Heisenberg indeterminacy relationship principles. The principle establishes the impossibility of certain pairs of observable and complementary physical quantities being known with arbitrary precision. If several identical copies of a system are prepared in a given state, such as an atom, the measurements of position and amount of movement will vary according to a certain probability distribution characteristic of the quantum state of the system. The measurements of the observable object will suffer standard deviation Δx of the position and the moment Δp. They then verify the principle of indeterminacy that is expressed mathematically as:

$\Delta x \cdot \Delta p \geq \frac{h}{2\pi}$

Here

h = Planck's constant

We have then,

$\Delta p \geq \frac{h}{2\pi \Delta x}$

Replacing our values we have that

$\Delta p \geq \frac{1.0546*10^{-34}J\cdot s}{2(100*10^{-12}m)}$

Note that the value substituted is the direct division between the Planck constant and the constant $\pi$

$\Delta p \geq 5.3*10^{-25} kg\cdot m \cdot s^{-1}$

Now the speed will be taken from the momentum expression that defines as

$\Delta p = mv \rightarrow v = \frac{\Delta p}{m}$

Remember that the momentum is the product between mass and velocity, replacing we will have

$v = \frac{\Delta p}{m}$

$v = \frac{5.3*10^{-25}kg \cdot m \cdot s^{-1}}{9.11*10^{-31}kg}$

$v = 5.8*10^5m/s$