# Calculating the most probable radius for an electron of a hydrogen atom in the ground state

This link describes a method for determining the most probable radius of an electron for a Hydrogen atom in the ground state.

It states that :

The radial probability density for the hydrogen ground state is obtained by multiplying the square of the wavefunction by a spherical shell volume element.

When I went to solve this problem myself I multiplied the square of the given wavefunction by the volume of a sphere, which gave me the wrong answer as I know it should be the Bohr radius.

When I thought about this problem it seemed reasonable to multiply it by the volume of a sphere rather than the surface area of a sphere (my gut feeling). The link doesn't really explain to me why it uses a surface area and not a volume of a sphere and I would like some help in getting an intuition for why the surface area is used and gives the right answer and the volume of a sphere does not.

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Note: ChocoPouce's answer is the same as mine but is more mathematical.

You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this distribution is in at the $0.1\mathrm{nm}$ shell? We won't exactly $0.1\mathrm{nm}$, so we take a thin shell: how much is between $0.1$ and $0.1+ε\mathrm{nm}$? Amount $=$ (average prob. density in shell)*(volume of shell). Since the shell is so thin ($\varepsilon$ is very small), the density will be almost constant and the shell's volume is given by $\mathrm{area}\times\mathrm{thickness} = 4\pi*(0.1nm^2)*\varepsilon nm$.

Thus the probability is $4\pi\rho(r)r^2\varepsilon$, where $\rho$ only depends on $r$ since it is spherically symmetric. But $\varepsilon$ is an arbitrary "small" thickness we defined, and it is best to divide by $\varepsilon$ to get the probability per unit radius, which is called the (radial) probability density $4\pi\rho(r)r^2$. The most likely radius (which is different from the average radius) maximizes this function.

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Ok so I understand some of what you are both saying, that the probability density * the volume enclosed by two spheres allows me to get to the answer. But it still doesn't make intuitive sense to me that the volume between two spheres (spherical shell) is used and not the volume of a sphere from the centre to some arbitrary distance r. I can remember to do this in future, but if the distribution is spherically symmetric why not use a volume element of a sphere instead of a spherical shell? –  Aesir Apr 7 '14 at 16:51
We use area because we want the probability that r is exactly 0.1 nm (technically probability density), not the probability that r is less than 0.1 nm. –  Kevin Kostlan Apr 7 '14 at 18:36
Thanks this is what made it click for me. –  Aesir Apr 7 '14 at 18:59

The link uses a spherical shell element which is $4\pi r^2 \mathrm{d}r$ and has the dimension of a volume ($r^2$ is a surface and $\mathrm{d}r$ is a length.

The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as $r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi$. There is no surface element used in the calculation. To see how the spherical shell volume element is related to spherical symmetry you may see that : $$\int_0^R \mathrm{d}r \int_0^{\pi} \mathrm{d}\theta \int_0^{2 \pi} \mathrm{d}\varphi r^2 = 4 \pi \int_0^R r^2 \mathrm{d}r$$

The spherical shell volume element already contains the integration over angles.

To perform those calculation one has to multiply the wavefunction by an volume element and not by the total volume of a sphere. The definition of the wavefunction states that it is a function $\psi(r,t)$ such as $|\psi(r,t)|^2 \mathrm{d}V$ is the probability to find the particle in a volume element $\mathrm{d}V$ around the point denoted by $r$ (in spherical coordinates on this example). Multiplying the wavefunction by a volume (to get an amplitude of probability) makes sense only if the volume is a small one.

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Thanks but I am still confused. I have the following formula $P(r)=\int^{2\pi}_0 \int^{\pi}_0|\psi(r)|^2r^2\sin\theta\mathrm{d}\theta\mathrm{d}\varphi=|\psi(r)|^‌​2 4\pi r^2$. Which I thought to be the probability of finding the electron on a radius r and also the surface area of a sphere. Then I can proceed to find the maximum by computing $\frac{\mathrm{d}P}{\mathrm{d}r}=0$. If I accept that it is instead a volume of a spherical shell element (which I think I get now) I still do not get why a spherical shell element is used and not the volume element of a sphere itself? –  Aesir Apr 7 '14 at 14:52