GR9277 #30



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bucky0 20071004 17:11:46  For positronium, I always assume that 6.8 eV is the binding energy. It's quick and right.   agaliarept 20061125 16:00:09  How can Z=1 when there is no Proton or Neutoron in positronium?
Jeremy 20071030 18:11:20 
In this formula, is used to represent the number of positive charges in the nucleus. In typical cases, this represents the number of protons. Here however, it is a positron that contributes the charge.

  tera 20060821 04:58:39  Positron has 1/1800 the mass of hydrogen i think and i can not see the formula tha says that energey is proportional to the mass. I just can not understand the correct answer
caffeinated 20080409 10:13:33 
The Rydberg constant is E/hc =911.27E1 angstoms. For H, divide by 1+m/M which is the reduced mass. In this case, M=m, since the nucleus is a positron, it has the same mass as the electron. Thus, the ratio of energy is 1/2.

dcan 20080410 10:00:49 
Oops. I left out . The Rydberg constant is . I sometimes forget about Z because it's usually H with Z=1.

Almno10 20101112 14:02:03 
The reduced mass of hydrogen atom is about me, since mu = (mp*me)/(mp + me). Do this for positronium, and see that the reduced mass is (1/2)me

Marian 20160105 08:59:41 
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  astro_allison 20051124 04:08:02  are you sure z=2 here? the math doesn't work. plus it doesn't make sense that positronium, with a smaller reduced mass than helium (z=1) would have a higher z. shouldn't z=1?
yosun 20051126 01:34:59 
astro_allison: thanks for the typoalert; it has been corrected.

 

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