# momentum in calculation of Bohr's radius

Some usual calculations of the Bohr's radius (see 2.5 at Feynman's here, or this text here) starts by defining a radius $$a$$ ( most probable radius ? average radius ? ). Next, from Heisenberg's uncertainty principle:

$$\Delta p \Delta a \ge h/2$$

it is said:

$$p = h/a$$.

where $$p$$ is the electron momentum.

Remainder of the calculation is expression of the total electron energy and minimization.

This step from uncertainty to momentum is confusing to me: we pass from inequality to equality, from margins ($$\Delta$$) to concrete values of $$p$$ and $$a$$.

Any hint to understand this step? Thanks.

This is another similar example, from "Quantum Mechanics", Nouredine Zettili, example 1.6:

Estimate the uncertainty in the position of (a) a neutron moving at $$5 \cdot 10^6 m/s$$ ... Solution: Using (1.57), we can write the position uncertainty as $$\Delta x = \frac{\hbar}{2 \Delta p} = \frac{\hbar}{2 m v } = ...$$

again, $$\Delta p$$, an interval, is converted to the absolute value $$p=mv$$.

If something is in an interval $$p \in [p_{min},p_{max}]$$ we can say $$p=p_{avg}\pm \Delta p$$ where $$p_{avg}=\frac{p_{max}+p_{min}}{2}$$ and $$\Delta p=\frac{p_{max}-p_{min}}{2}$$, but these texts seems to assume $$p_{min}=0$$ and $$\frac{p_{max}}{2}=mv=\Delta p$$ or something similar.

In other words, when in this example we give a known value for the speed, we are fixing the momentum without any indetermination, thus, uncertainly in position should be infinite.

• From the Feynman source: "We now consider another application of the uncertainty relation, Eq. (2.3). It must not be taken too seriously; the idea is right but the analysis is not very accurate." – jacob1729 Jan 27 '19 at 19:57

• if we are using the uncertainty relation, $$x,p$$ do not have single simultaneous values and so it makes no sense to define $$p=h/a$$;
• the uncertainty relation and the virial principle themselves do not actually fix the size of the atom, the atom can be as large as we want - great size $$a$$ corresponds to a high excitation number $$n$$.
For example, we can assume that minimum size of atom is achieved when the expected average of its energy is the minimal possible value, and then try to use HUP and the virial principle to derive some estimate of atom size (defined as $$\sqrt{\langle x^2\rangle}$$). Similar procedure is known in the general theory of electron shells of an atom, where it is able to provide and estimation of the ground state energy and the corresponding psi function.
• Maybe it comes from the fact that $p=h/\lambda$ and $\lambda=2\pi a$ so we would have $p=h/2\pi a$ or $pa=\hbar$ and its bigger then the limit given by the uncertanity principle. – Layla Jan 28 '19 at 4:37