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GR0177 #93
Problem
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Quantum Mechanics$\Rightarrow$}Most probable value

Recall the definition of probability, $P=\int |\psi|^2dV=\int|\psi|^2 4\pi r^2 dr$.
(The radial probability distribution $P_r$ is related to the probability $P$ by $P=\int P_r dr$.)
The most probable value is given by the maximum of the probability distribution. Taking the derivative with respect to r, one has this condition for a maximum (the second derivative shows that it's concave up)

$dP_r/dr = 4\pi r^2 d|\psi|^2/dr + 8\pi r |\psi|^2=0$.

The problem gives $\psi = \frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0}$. Thus, $|\psi|^2=\frac{1}{\pi a^3_0}e^{-2r/a_0}$ and $d|\psi|^2/dr = \frac{-2}{\pi a^3_0 a_0}e^{-2r/a_0}$.

Plug this into the expression for $dP/dr$, solve for $r$ to find that the most probable distance is just the Bohr radius, as in choice (C).

Alternate Solutions
 radicaltyro2006-10-28 23:38:14 One can also note that this is more or less the definition of the bohr radius.Reply to this comment
UNKNOWNUMBER
2018-10-20 23:28:34
I think this is one of those problems ETS expects you to know the answer to by heart after taking a quantum class.
bhaynor
2009-10-03 12:30:22
I get C. It asks for the most probable value of r, not the expectation value of r. The most probable value of r occurs when Psi is maximized, at r = 0.
 bhaynor2009-10-03 12:32:07 Ignore my last comment. I forgot about r^2.
anmuhich
2009-03-17 13:09:47
You can also remember that in general lower energy states are more probable and that the lowest energy state of hydrogen is the ground state at the bohr radius.
Imperate
2008-10-04 14:52:04
An interesting note on this question, is that the most probable value of r for the ground state wavefunction, is NOT equal to the expectation value.rnrnThe expectation value is actually 1.5a_o.rnrnhttp://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html#c3rnrnshows nice solutions to both of these related problemsrnrn
2006-10-28 23:38:14
One can also note that this is more or less the definition of the bohr radius.
 mitama2008-08-09 17:56:58 Depending on the wavefunction, this may or may not be true (luckily, in this case it is). It's probably best to go through the work just to be sure.
 syrock2009-10-08 16:50:41 yeah, hydrogen.... bohr.... come on its gatta be a0
 Almno102010-11-10 19:56:42 I don't think this is the definition of the bohr radius. The bohr model is classical. The "come one, its bohr" argument is correct. The bohr model is very good despite its crappy, classical derivation.
 walczyk2011-04-07 20:55:37 Almno10: The brilliant thing about bohr is that he reached this result with his own proto-quantum mechanics, and thus the radius is named after him. go read a book
 walczyk2011-04-07 20:56:55 nvm, read it wrong and didn't mean to post a dick comment :/
jax
2005-11-30 19:03:20
Why does dV become $4 \pi r^2$? Isn't that the surface area ?
 yosun2005-11-30 20:19:06 jax: thanks for the typo alert; the typo's been fixed: $dV=4\pi r^2 dr$. (The radial probability distribution $P_r$ is related to the probability $P$ by $P=\int P_r dr$.)

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