# Most probable value of r in ground state of hydrogen

1. Oct 26, 2004

### yxgao

What is the most probable value of r for the grounds tate of hydrogen, and why? Is it just the Bohr radius?

2. Oct 26, 2004

### Tom Mattson

Staff Emeritus
Yes, it is the Bohr radius (which I'll call 'a'). The ground state wavefunction is:

ψ1s(r)=(1/π1/2a3/2)e-r/a.

The probability density is |ψ1s(r)|2, which is:

ρ1s(r)=|ψ1s(r)|2=(1/πa3)e-2r/a.

But that function is not going to give you the most probable radius. You have to take into account the fact that ρ1s is in spherical coordinates, whose volume element is:

dV=r2sin(φ)dr dθ dφ.

So, when you integrate ρ1s over all space, it gets multiplied by r2. Furthermore, since ψ1s is spherically symmetric, you can integrate over θ and φ to get what is called the radial probability density P1s(r):

P1s(r)=(4/a3)r2e-2r/a.

If you optimize this function, you will find that it has a relative maximum at r=a, the Bohr radius.

Last edited: Oct 26, 2004
3. Oct 26, 2004

### yxgao

Thank you so much!! That was very helpful!

4. Oct 26, 2004

### Orion1

Particle Probability...

Quantum Hydrogen

Hydrogen wave function:
$$\psi_{1s} = \frac{1}{\sqrt{\pi r_0^3}} e^{-r/r_0}$$

$$r_0 = \frac{\hbar}{\alpha M_e c}$$

Probability density:
$$|\psi_{1s}|^2 = \left( \frac{1}{\pi r_0^3} \right) e^{-2r/r_0}$$

$$P(r)dr = |\psi|^2 dV$$

$$dV = 4 \pi r^2 dr$$

$$P(r)dr = |\psi|^2 4 \pi r^2 dr$$

Beta cloud probability:
$$P(r) = 4 \pi r^2 |\psi|^2$$

Hydrogen probability density function:
$$P_{1s}(r) = \left( \frac{4r^2}{r_0^3} \right) e^{-2r/r_0}$$

The most probable value of $$r$$ corresponds to the peak of the plot of $$P(r)$$ versus $$r$$. The slope of the curve at this point is zero. To evaluate the most probable value of $$r$$ is by setting $$dP/dr = 0$$ and solving for $$r$$:

$$\frac{dP}{dr} = 0$$

$$\frac{dP}{dr} = \frac{d}{dr} \left[ \left( \frac{4r^2}{r_0^3} \right) e^{-2r/r_0} \right] = 0$$

Derivative operation and simplification:
$$e^{-2r/r_0} \frac{d}{dr} (r^2) + r^2 \frac{d}{dr} (e^{-2r/r_0}) = 0$$

$$2re^{-2r/r_0} + r^2(-2/r_0)e^{-2r/r_0} = 0$$

$$2re^{-2r/r_0} - (2r^2/r_0)e^{-2r/r_0} = 0$$

$$2r[1 - (r/r_0)]e^{-2r/r_0} = 0$$

Expression satisfied if:
$$1 - \frac{r}{r_0} = 0$$

Therefore:
$$\boxed{r = r_0}$$

Last edited: Oct 26, 2004