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## Main Question or Discussion Point

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

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

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Tom Mattson

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

ψ_{1s}(r)=(1/π^{1/2}a^{3/2})e^{-r/a}.

The probability density is |ψ_{1s}(r)|^{2}, which is:

ρ_{1s}(r)=|ψ_{1s}(r)|^{2}=(1/πa^{3})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=r^{2}sin(φ)dr dθ dφ.

So, when you integrate ρ_{1s} over all space, it gets multiplied by r^{2}. 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/a^{3})r^{2}e^{-2r/a}.

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

ψ

The probability density is |ψ

ρ

But that function is not going to give you the most probable radius. You have to take into account the fact that ρ

dV=r

So, when you integrate ρ

P

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

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Thank you so much!! That was very helpful!

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Quantum Hydrogen

Hydrogen wave function:

[tex]\psi_{1s} = \frac{1}{\sqrt{\pi r_0^3}} e^{-r/r_0}[/tex]

Bohr radius:

[tex]r_0 = \frac{\hbar}{\alpha M_e c}[/tex]

Probability density:

[tex]|\psi_{1s}|^2 = \left( \frac{1}{\pi r_0^3} \right) e^{-2r/r_0}[/tex]

[tex]P(r)dr = |\psi|^2 dV[/tex]

[tex]dV = 4 \pi r^2 dr[/tex]

[tex]P(r)dr = |\psi|^2 4 \pi r^2 dr[/tex]

Beta cloud probability:

[tex]P(r) = 4 \pi r^2 |\psi|^2[/tex]

Hydrogen probability density function:

[tex]P_{1s}(r) = \left( \frac{4r^2}{r_0^3} \right) e^{-2r/r_0}[/tex]

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

[tex]\frac{dP}{dr} = 0[/tex]

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

Derivative operation and simplification:

[tex]e^{-2r/r_0} \frac{d}{dr} (r^2) + r^2 \frac{d}{dr} (e^{-2r/r_0}) = 0[/tex]

[tex]2re^{-2r/r_0} + r^2(-2/r_0)e^{-2r/r_0} = 0[/tex]

[tex]2re^{-2r/r_0} - (2r^2/r_0)e^{-2r/r_0} = 0[/tex]

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

Expression satisfied if:

[tex]1 - \frac{r}{r_0} = 0[/tex]

Therefore:

[tex]\boxed{r = r_0}[/tex]

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