GR0177 #31



Alternate Solutions 
FutureDrSteve 20111107 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= 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).  

Comments 
bill92 20141027 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 20111107 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= 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 20071031 17:48:57  Why is the reduced mass m/2 ?
nick1234 20071102 17:16:11 
For the hydrogen atom, dominates the denominator, and becomes
For positronium, , so

  rinnie 20070326 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.
FutureDrSteve 20111107 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= 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).

 

Post A Comment! 
You are replying to:
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= 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).

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .

type this... 
to get... 
$\int_0^\infty$ 

$\partial$ 

$\Rightarrow$ 

$\ddot{x},\dot{x}$ 

$\sqrt{z}$ 

$\langle my \rangle$ 

$\left( abacadabra \right)_{me}$ 

$\vec{E}$ 

$\frac{a}{b}$ 





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

