# QM approximation (electron within nucleus, Griffiths 4.45b

• Saraphim
In summary, the probability that an electron in the ground state of hydrogen will be found inside the nucleus is approximately 4/3.
Saraphim

## Homework Statement

What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus?

a) First calculate the exact answer, assuming the wave function $\psi(r,\theta,\phi) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$ is correct all the way down to r=0. Let b be the radius of the nucleus

b) Expand your result as a power series in the small number $\epsilon = 2b/a$ and show that the lowest-order term is the cubic: $P \approx (4/3)(b/a)^3$. This should be a suitable appoximation, provided that b << a, which it is.

## The Attempt at a Solution

Using partial integration a few times, I've gotten an answer for a) which I think is sufficient in order to solve b). $P (r<b) = 1-e^{-2b/a}(2b^2/a^2 + 2b/a + 1)$.

However, I'm completely lost on b). I'm looking for pointers here on how to even attack the problem. I can rewrite P so that I get $P = 1 - e^{-\epsilon}(\epsilon/2 + \epsilon + 1)$ but from there I'm stumped. I think the problem is that I've mostly forgotten how to do Taylor expansion, and a quick lookup does not help me in attacking this specific problem.

Hi Saraphim!
Saraphim said:
… show that the lowest-order term is the cubic: $P \approx (4/3)(b/a)^3$

$P (r<b) = 1-e^{-2b/a}(2b^2/a^2 + 2b/a + 1)$.

use the definition of ex (which is also its taylor series) …

ex = 1 + x + x2/2 + x3/6 + …​

then P = 1 - e-2ε(1 + 2ε + 2ε2)

= 1 - (1 - 2ε + 4e2/2 - 8ε3/6 + …)(1 + 2ε + 2ε2)

= … ?

Thanks for your response! I have a few problems, still. First, I feel the Taylor expansion should be $1 - \epsilon + \epsilon^2/2 - \epsilon^3/6 +\ ...$. Isn't that right?

Second, I feel that I should be able to realize that the terms lower than the cubic somehow get canceled out, but I really don't see it. :(

Good morning!
Saraphim said:
Thanks for your response! I have a few problems, still. First, I feel the Taylor expansion should be $1 - \epsilon + \epsilon^2/2 - \epsilon^3/6 +\ ...$. Isn't that right?

that would be correct for e

the question gives you e-2ε
Second, I feel that I should be able to realize that the terms lower than the cubic somehow get canceled out, but I really don't see it. :(

should cancel out now!

But how is that so? In the unexpanded P without substituting in $\epsilon = 2b/a$, I have $e^{-2b/a}$.

oh, I'm sorry, i was using ε = b/a

with ε = 2b/a,

then P = 1 - e(1 + ε + ε2/2)

= 1 - (1 - ε + e2/2 - ε3/6 + …)(1 + ε + ε2/2)

I think I understood what I was missing. I needed to set all terms in the product with exponents higher or equal to 4 to zero. Approximation is not my string suite. Thanks for the help!

## What is the QM approximation for the electron within the nucleus?

The QM approximation, also known as the quantum mechanical approximation, is a method used to simplify the calculation of the electron wave function within the nucleus. It assumes that the potential energy inside the nucleus is much greater than the kinetic energy of the electron, and therefore the electron's motion can be approximated as simple harmonic motion.

## How is the QM approximation used in the context of Griffiths 4.45b?

In Griffiths 4.45b, the QM approximation is used to calculate the ground state energy of a hydrogen-like atom, where the electron is confined within the nucleus. The approximation allows for a simpler and more tractable solution to the Schrödinger equation, which describes the behavior of quantum particles.

## What are the limitations of the QM approximation?

The QM approximation is only valid for systems with a large potential energy difference between the electron and the nucleus, such as in the case of a hydrogen-like atom. It also does not take into account any electron-electron interactions, which can become significant in multi-electron atoms.

## How accurate is the QM approximation?

The accuracy of the QM approximation depends on the specific system being studied. In the case of the hydrogen atom, the approximation yields results that are within a few percent of the exact solution. However, for more complex systems, the accuracy may be lower and other methods may be necessary.

## Are there any other approximations used in the study of quantum mechanics?

Yes, there are many other approximations used in quantum mechanics, such as the Born-Oppenheimer approximation, the Hartree-Fock approximation, and the time-independent and time-dependent perturbation theory. These approximations are used to simplify the calculations and make them more manageable, but they may introduce some level of error in the results.

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