[time-nuts] Disciplining Rubidium

[SAIDJACK at aol.com]

Hi Bruce,
 
wow, haven't done that math since about 1987.. I remember now.
 
How about Cs? Seems more aggressive.
 
bye,
Said
 
 
In a message dated 4/23/2008 23:52:31 Pacific Daylight Time,  
bruce.griffiths at xtra.co.nz writes:

SAIDJACK at aol.com wrote:
> Hi Bruce,
>  
> last  chemistry/physics class is a while back :) I guess a half life of 50  
 
> Billion years means it's not really radiating much?
>   
> No problem with Cesium then either, I guess the radiation levels must  be  
> really really low?
>  
> thanks,
>  Said
>   
Said

Calculating the number of atoms that  decay per second in a particular 
sample is almost trivial:
Divide the  mass of the sample (in grams) by the atomic weight, multiply 
the result by  Avogadro's number and then divide by the number of seconds 
in 5E10  years.

Result for a 10gm sample of Rb87

No of atoms present ~  6.02E23 x 10/85 ~ 7E22
No of seconds in 5E10 years ~ 1.5E18
Thus number  atoms in the sample decaying per second ~47,000.
Equivalent electron  current ~ 7.6fA.

One saving grace is that beta particles (electrons)  are essentially 
stopped by a piece of paper, just dont eat the  stuff.

Bruce

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