A proton is confined to a nucleus whose diameter is 5.5*10^-15m. If this distance is considered to be the uncertainty in the position of the proton, what is the minimum uncertainty in its momentum?

I don't really get the uncertainty concept, and how can I do this question?Please help!THANKS A LOT!

To understand uncertainty, you need to know about Heisenberg's uncertainty principle. According to this principle, it is impossible to simultaneously know the exact position and momentum of a quantum particle. The uncertainty in position multiplied by the uncertainty in momentum is always greater than or equal to a certain constant value known as Planck's constant (h). Mathematically, it can be expressed as:

Δx * Δp >= h/4π

Here, Δx represents the uncertainty in position, and Δp represents the uncertainty in momentum.

In this question, you are given the uncertainty in position of the proton, which is the diameter of the nucleus (5.5*10^-15m). You need to find the minimum uncertainty in its momentum.

To find the minimum uncertainty in momentum (Δp), you can rearrange the uncertainty principle equation:

Δp >= h/4πΔx

First, let's determine the value of Planck's constant (h), which is approximately 6.63 × 10^-34 J·s.

Now plug in the given values:

Δx = 5.5 × 10^-15 m
h = 6.63 × 10^-34 J·s

Δp >= (6.63 × 10^-34 J·s) / (4π × 5.5 × 10^-15 m)

Perform the calculations:

Δp >= (1.05 × 10^-33 J·s)/(1.74 × 10^-14 m)

Simplifying further:

Δp >= (1.05 × 10^-19 J·s·m^-2)

So, the minimum uncertainty in the momentum of the proton is approximately 1.05 × 10^-19 J·s·m^-2.

Remember that this is the minimum uncertainty; the actual uncertainty may be higher than this value due to other factors.