Calculating the location where the Radial probability density maximum at n=2 l=0 state of hydrogen?

Question

Also in association with this question any recommnedation where I can find how to calculate from first principles the expectation value of the radial coordinate r in this state??

Answer 1

This is covered in most quantum mechanics textbooks. I'm looking at the appendix to Albert Messiah's Quantum Mechanics Vol I. The wavefunction for n=2,l=0 is:

(1/(a^3/2)) * 1/sqrt(8pi) * (1-r/2a) * exp(-r/2a)

where a is the Bohr radius. Let's use units in which a=1. The probability density is:

P(x) = 1/(8pi) * (1-r/2)^2 * exp(-r)

The radial probability density is the above times the area of the sphere:

R(r) = 4pi * r^2 * P(x) = 1/2 * r^2(1-r/2)^2 * exp(-r)

If you want the radius of the location in space at which P(x) is maximal solve:

dP(r)/dr = 0 (a quadratic equation)

If you want the radius at which the radial probability density is maximal solve:

dR(r)/dr = 0 (a cubic equation)

R(r) has two relative maxima. I think the one farther out is bigger. The expectation value of the radius is:

= \int_0^oo R(r) r dr

straightforward using \int_0^oo r^n e^(-r) dr = n! etc.. The answer should be 6a.

Answer 2

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