Why are we able to use the uncertainty principle in quantum mechanics to calculate momentum/position?

Question

I was asked the following question in my textbook:

The nitrogen molecule consists of two nitrogen atoms joined by a covalent bond with length approximately 100pm. What is the approximate kinetic energy of the covalently bonded electrons?

The process to solving this is to use the uncertainty relation
$$Delta p Delta x approx hbar$$
and insert 100pm into $Delta x$ to solve for $Delta p$. Then you use
$$E = frac{p^2}{2m_e}$$
where $m_e$ is the mass of an electron to find the kinetic energy.
I understand that $Delta x$ is around 100pm because that's the distance between the nitrogen atoms, so it's generally the range in which the electron can roam. But then we use that $Delta x$ to find $Delta p$ and treat $Delta p$ as though it were just momentum $p$. Why can we say, in this instance, that $Delta p = p$?

Superfast Jellyfish · Accepted Answer

In this case, under the Born approximation, nuclei at rest compared to the electron motion. Thus effectively the electrons are bound within a region by the nuclei. This means that the average momentum of the electrons must be zero.
Thus in this case, the spread in momentum is itself roughly, the momentum. $Delta papprox p$

Why are we able to use the uncertainty principle in quantum mechanics to calculate momentum/position?

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