While solving Schrödinger equation for Hydrogen atom we make a scale transformation for radial variable ($r=\frac{ax}{Z}$; where $a=$ Bohr radius, $x=$ dimensionless variable and $Z=$ atomic number), this turns out to be a very good scale transformation. But my question is how do we know value of Bohr radius in advance, before solving Schrödinger equation? Do we just use Bohr radius that we got from Bohr theory? If we do use Bohr radius from Bohr theory then why is so because it is a classical theory?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ It pops out from the Buckingham-$\pi$ theorem. $\endgroup$– user14717Commented Jun 8, 2020 at 1:05
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
this turns out to be a very good scale transformation.
We don't really care that it's a good or bad choice. We start by choosing a scale factor with just a symbol to move to dimensionless coordinates and follow the math through using it. The physical problem is solved no matter what we choose (as long as it's not silly like zero).
We can at any point in going through the math (on this or any problem where we apply this technique) choose a value for the scale factor that is convenient and makes doing the math simpler.
But my question is how do we know value of Bohr radius in advance, before solving Schrödinger equation?
We don't. We solve the problem and then we label the value as the Bohr radius. That's what happened when Bohr's original model was published - it did not start out being called the "Bohr radius, it became that by convention.
For some problems the scale factor we end up with for mathematical convenience has a straight-forward physical interpretation (a radius, an orbital period, a reduced mass, etc.), but sometimes it's just a constant in the equation because that's how the mathematics works out.
answered Jun 7, 2020 at 16:02
$\begingroup$
$\endgroup$
4
Usually when transforming into dimensionless variables one looks at the relevant constants in the problem. For hydrogen atom we have the electron charge $e$, electron mass $m_e$, Plank constant $\hbar$, permittivity of free space $\epsilon_0$
Then one does dimensional analysis to make scales based on the above constants. And the expression for length scale turns out to be $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}}$$ The extra $4\pi$ is part of the permittivity. And this is exactly the Bohr radius.
answered Jun 7, 2020 at 15:40
-
$\begingroup$ Yes, I thought of that but relevant constants in the problem which have dimensions are Epsilon, h, e, Me so dimensional analysis will give a distance without a factor of 4pi.we say that there should be 4pi because we know actual value of 4pi?right? $\endgroup$– BarryCommented Jun 7, 2020 at 16:22
-
$\begingroup$ Yes. To get the exact Bohr radius it is relevant. But you get the order of magnitude of the lengthscale to a good estimate by this. Basically what’s the domain of validity of the theory. Remaining (exact) value is determined out of the calculation as @StephenG pointed out in his answer. $\endgroup$ Commented Jun 7, 2020 at 16:30
-
1$\begingroup$ In the formula for $a_0$ it should be $m_e$, not $m_e^2$. $\endgroup$ Commented Jun 7, 2020 at 16:48
-
$\begingroup$ Right! Thanks @Thomas $\endgroup$ Commented Jun 7, 2020 at 17:00