« first day (1457 days earlier)   

2:19 AM
@ChrisWhite Brookhaven states they are both pure electron capture
@ChrisWhite my homework says the parent is pure electron capture and the daughter is a mixture
(yet Brookhaven lists 100% electron capture for the daughter, but gives energies for the Beta+ emitted)
 
2:36 AM
There does indeed appear to be conflicting information
Could be some poorly entered/formatted data in someone's database.
The other thing that comes to mind is that electron capture and beta+ emission are basically the same -- a nucleus susceptible to one is susceptible to the other. But EC depends on there actually being electrons. So if you fully ionize the atom, you can wait around for the rare positron, since nothing else can happen anyway.
 
 
2 hours later…
4:14 AM
@ChrisWhite That was my initial perception and multi-modal decay constants sum nicely to give a single effective half life anyway
 
 
2 hours later…
6:31 AM
@IceBoy you here ? :)
 
6:51 AM
@TheArtist yep :)
 
@IceBoy If your free can I ask you a physics question? :)
 
AskAway
 
@IceBoy this is the question :) I successfully showed the moment of intertia is that value....the problem is in my second part......let me explain :)
Since C is the center of mass....by symmetry....i drew it like this ....where in the first step I found the distance OC , and then I drew OC along a Line.....and since this will go like a circle, with radius OC....I thought the greatest $\omega$ would be at the lowest point....when taking conservation of energy :) and then I did :
You can also see the "zero PE line" I have drawn :) ....
Then I applied that the total potential energy turns to kinetic energy at the bottom which is :
Mgh at top changes to $\frac{1}{2} I \omega^2$ at the bottom.....I get a value for $w=12.8$ :
@IceBoy and I did the calculations from start about 3 times to make sure it's not an arithmetic error :) any idea what I have done wrong? :)
 
no idea, perhaps @JamalS or @hwlau can help
 
Correct answer is $6.75 rad s^{-1}$
@IceBoy okay :D :)
@IceBoy are you a university student? :) if so which univ? :)
@IceBoy oh cool :) ok :)
 
 
5 hours later…
11:58 AM
@TheArtist I think the problem is that you are using the moment of inertia about the center; you need to use the parallel axis theorem to compute the moment of inertia about C.
 
12:09 PM
I assume you know how to take it from here...
I just checked - and that additional factor gets one to the right answer (whilst otherwise using your method - just with the corrected I).
 
12:30 PM
@Floris thanks for helping him out :)
 
@IceBoy no problem. I just stumbled in here and the answer just jumped out at me.
 
@Floris btw, you have a typo in your profile description " I believe many problems can be solved my by making a careful sketch."
 
@GBeau - your second decay constant (the slower one) can be found by the asymptote of the combined decay curve (where the parent has essentially decayed completely so all decays you see are from the daughter) - just as the first is given by the initial slope
@IceBoy - thanks for noticing. I will fix it.
 
@Floris Oh :) Thank you very much Floris :)
@IceBoy ^_^ :)
 
12:57 PM
Welcome. It's an easy thing to overlook - and so much easier to spot with fresh eyes...
 
1:13 PM
:)
Are you a physicist ? :)
@Floris you have such a large reputation score :)
 
Yes I am a physicist. Yes, I spend too much time answering questions...
But now I have to go and do some "real" work... TTYL
 
Wow that's so cool :) ttyl :) tc :)
 
 
2 hours later…
2:52 PM
Mods: What would you prefer I do with a link-only answer like this? Flag as NAA or vote to delete?
 
If it matters any, I've already flagged it as NAA
It popped up in the first post queue
 
Another flag from me should be there, too. I edited it from the LQ queue, forgetting that that takes it out of the queue, and flagged it again...
 
Hmm, the question is in the queue for me, but not the answer. Nor is it in any other queue at the moment. I suppose there's some lag.
 
 
5 hours later…
7:25 PM
In applications where it is natural to use the angular frequency (i.e. where the frequency is expressed in terms of radians per second instead of rotations per second or Hertz) it is often useful to absorb a factor of 2π into the Planck constant. The resulting constant is called the reduced Planck constant or Dirac constant. It is equal to the Planck constant divided by 2π, and is denoted ħ (pronounced "h-bar") Source: Wikipedia. Note the hyphen between "h" and "bar"?
The h Bar note the lack there of.
 
 
1 hour later…
8:53 PM
@IceBoy Is it possible that you have missed the joke?
Let me buy you a drink. You'll think more clearly then.
 
Are we at a bar? ;)
4
 
I don't know about you, but I am.
 
lol :)
 
@dmckee Ahh...icic :D)
too much literal math
 
::raises a neat whiskey:: Cheers!
 
8:56 PM
mmmm....whiskey.....
I haven't had any since February when my supply ran dry
 
9:27 PM
@dmckee cheers!
 

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