GREPhysics.NET: GR8677 #99

GR8677 #99

Problem

GREPhysics.NET Official Solution

Atomic $\Rightarrow$ }Positronium

Interestingly, this is almost an exact repeat of Exam 9677 Prob 12. The reasoning's the same:

The positronium atom involves a positron-electron combination instead of the usual proton-electron combo for the H atom. Charge remains the same, and thus one can approximate its eigenvalue by changing the mass of the Rydberg energy (recall that the ground state of the Hydrogen atom is 1 Rydberg).

Recall the reduced mass $\mu=\frac{m_1 m_2}{m_1+m_2}$ , where for identical masses, one obtains $\mu = m/2$ . The Rydberg in the regular Hydrogen energy eigenvalue formula $E=R\left(1/n_f^2 - 1/n_i^2 \right)$ is $R\propto \mu$ , where $\mu \rightarrow \mu/2$ . Substitute in the new value of the reduced mass to get $E\approx R/2$ . $R=-13.6$ eV, and thus $E\approx -6.8$ eV.

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

Geez
2015-10-22 23:02:29

It\\\\\\'s 13.6 ev with ~infinitely massive proton -> no proton KE. If positron can move, it will share same pool of energy with electron. The coulomb potential is the same as for hydrogen because the charges are the same as with hydrogen, so they each get (1/2) the total energy since they are same mass. -> (1/2)13.6 eV = answer B, as given in booklet.

Reply to this comment

cz
2012-11-04 22:22:31

Not really a great way of doing problems. Your method will give the answer here. The reason that happens is because in the actual formula for the ground state energy is proportional to $m\cdote^4\cdotZ^2$
Here m is the reduced mass of the electron.

For the hydrogen atom it is just the same as electron mass (since the nucleus is so heavy).
For the positronium it is half of original mass, since the positron has the same mass as the electron.

Reply to this comment

TheDoctor
2012-08-22 12:22:17

Could you say that you know energy is proportional to mass (rest energy = $m*c^2$ ), thus it is proportional to reduced mass, and then make the reduced mass argument that you made about $\mu$ going to $\mu$ /2}?

cz
2012-11-04 22:25:23

Not really a great way of doing problems. Your method will give the answer here. The reason that happens is because in the actual formula for the ground state, energy is given by $m_ee^4Z^2$
Here m is the reduced mass.

For the hydrogen atom m is almost the same as the electron mass. (the nucleus is 1840 times heavier)

For positronium, m is half of the electron mass. (positronium is the same mass as the electron).

If in the formula $m^2$ or anything else like that appeared, then your line of argument would always fail.

cz
2012-11-04 22:26:26

Actually energy is just proportional to what I have posted. I left out the constants.

Reply to this comment

anum
2010-11-11 12:30:34

what i did was E directly proportional to m .m halved then energy halved.( by E i mean the ground state energy of hydrogen of which formula i have memorized)

Reply to this comment

Richard
2007-11-01 13:56:01

I think there is a positronium question akin to this one on all of the released phyiscs GRE's.
Hint I think.

wittensdog
2009-07-28 11:02:09

Everyone I have ever talked to who has taken the physics GRE has said there was a positronium question on their test.

I'm glad ETS gets to dictate what's important in physics. Apparently it's positronium. Yes. Positronium is the most important thing in physics. Ever.

I understand knowing how to use reduced mass is important, but still, it seems a little over the top to me.

jcsoldier11
2009-10-26 00:16:10

According to my friend who took it a couple weeks ago, there was a Positronium question on it in which he did basically the same thing as above.

mvgnzls
2011-09-14 19:00:13

postironium is important coming from a person working in the positron lab at UCR!!!

Reply to this comment

Blake7
2007-09-26 01:23:16

Yosun,

Don't you mean that 9677.12 is almost an exact repeat of 8677.99 ?

; )

Reply to this comment

Post a Comment

Comments are open for signed-in members.

A short Physics 101 question is required before each comment is published.

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