GR 8677927796770177 | # Login | Register

GR0177 #31
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Atomic$\Rightarrow$}Positronium

The positronium atom ground-state energy can be deduced from recalling the reduced mass. $\mu=m/2$, since one has a positron and electron orbiting each other, and thus the energy, which is related to mass, is half of that of Hydrogen. $E_{positronium}=-13.6/2 eV = -6.8eV$

The Bohr formula still applies, and thus $E=E_1 \left(1/n_f^2 - 1/n_i^2 \right)=8E_1/9$, which is choice (A).

Alternate Solutions
 FutureDrSteve2011-11-07 12:22:35 Oops, this was supposed to be a separate post, not a reply.... If you realize that the ground state energy of positronium is roughly -6.8eV (which I think you have to in order to solve this problem), you don't have to do any math. An electron transitioning down from n=$\infty$ would have initial potential energy of 0, and drop down to final energy -6.8eV, for a max energy change of 6.8eV, eliminating (C), (D), and (E). Since the electron is only falling down from n=3, it stands to reason that the photon would be less energetic than 6.8eV, leaving only answer (A).Reply to this comment
bill92
2014-10-27 13:17:41
I just took the exam and there was no positronium question! All my studying was for naught!

Just kidding, I think it went well. Thanks to Yosun for this very helpful site!
FutureDrSteve
2011-11-07 12:22:35
Oops, this was supposed to be a separate post, not a reply....

If you realize that the ground state energy of positronium is roughly -6.8eV (which I think you have to in order to solve this problem), you don't have to do any math. An electron transitioning down from n=$\infty$ would have initial potential energy of 0, and drop down to final energy -6.8eV, for a max energy change of 6.8eV, eliminating (C), (D), and (E). Since the electron is only falling down from n=3, it stands to reason that the photon would be less energetic than 6.8eV, leaving only answer (A).
Tommy Koulax
2007-10-31 17:48:57
Why is the reduced mass m/2 ?
 nick12342007-11-02 17:16:11 $\mu = \frac{m_{1} \times m_{2}}{m_{1} + m_{2}}$ For the hydrogen atom, $m_{proton}$ dominates the denominator, and $\mu$ becomes $\frac{m_{e} \times m_{p}}{m_{p}} = m_{e}$ For positronium, $m_{1} = m_{2} = m{e}$, so $\mu = \frac{m_{e}^{2}}{2 m_{e}} = \frac{m_{e}}{2}$
rinnie
2007-03-26 21:27:49
Hydrogen levels E3 - E1 = -1.5 + 13.6 = 12.1/2 = 6.05.
Energy level of positronium is half those of Hydrogen.
 FutureDrSteve2011-11-07 12:20:53 If you realize that the ground state energy of positronium is roughly -6.8eV (which I think you have to in order to solve this problem), you don't have to do any math. An electron transitioning down from n=$\infty$ would have initial potential energy of 0, and drop down to final energy -6.8eV, for a max energy change of 6.8eV, eliminating (C), (D), and (E). Since the electron is only falling down from n=3, it stands to reason that the photon would be less energetic than 6.8eV, leaving only answer (A).

Why is the reduced mass m/2 ?

This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

## 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...