Most probable value of r in ground state of hydrogen

[yxgao]

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

 

[quantumdude]

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

ψ_1s(r)=(1/π^1/2a^3/2)e^-r/a.

The probability density is |ψ_1s(r)|², which is:

ρ_1s(r)=|ψ_1s(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 ρ_1s is in spherical coordinates, whose volume element is:

dV=r²sin(φ)dr dθ dφ.

So, when you integrate ρ_1s over all space, it gets multiplied by r². Furthermore, since ψ_1s is spherically symmetric, you can integrate over θ and φ to get what is called the radial probability density P_1s(r):

P_1s(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.

 

[yxgao]

Thank you so much! That was very helpful!

 

[Orion1]

Particle Probability...


Quantum Hydrogen

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

Bohr radius:
$$r_0 = \frac{\hbar}{\alpha M_e c}$$

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