GREPhysics.NET: GR0177 #93

GR0177 #93

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

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

radicaltyro
2006-10-28 23:38:14

One can also note that this is more or less the definition of the bohr radius.

Reply to this comment

Comments

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.

Reply to this comment

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.

bhaynor
2009-10-03 12:32:07

Ignore my last comment. I forgot about r^2.

Reply to this comment

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.

Reply to this comment

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

Reply to this comment

radicaltyro
2006-10-28 23:38:14

One can also note that this is more or less the definition of the bohr radius.

mitama
2008-08-09 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.

syrock
2009-10-08 16:50:41

yeah, hydrogen.... bohr.... come on its gatta be a0

Almno10
2010-11-10 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.

walczyk
2011-04-07 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

walczyk
2011-04-07 20:56:55

nvm, read it wrong and didn't mean to post a dick comment :/

Reply to this comment

jax
2005-11-30 19:03:20

Why does dV become $4 \pi r^2$ ? Isn't that the surface area ?

yosun
2005-11-30 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 $P=\int P_r dr$ .)

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone