Calculation Error: Iodine-131 Half-Life & 30 Days

  • Thread starter Thread starter Ukitake Jyuushirou
  • Start date Start date
  • Tags Tags
    Half life Life
AI Thread Summary
Iodine-131, used for treating thyroid disorders, has a half-life of 8.04 days. After 30 days, calculations show that approximately 7.93% of the initial sample remains. The error in previous calculations stemmed from misapplying the mass number and misunderstanding the percentage calculation. The correct approach involves using the formula (1/2)^(30/8.04) to determine the remaining percentage. The discussion concludes with a clarification of the correct method for calculating the remaining percentage of Iodine-131.
Ukitake Jyuushirou
Messages
124
Reaction score
0
Iodine-131 is used in treatment of thyroid disorder. This isotope has a half life of 8.04 days, what percentage of an initial sample is left after 30 days?

having done, the initial calculations by taking 30 divided by 8.04 to find how many hl will have passed i take the number and i take the mass number divided by the amount of hl that has passed times 100 but my answers dun match the answer in the book.

where did i go wrong?
 
Last edited:
Physics news on Phys.org
No of HLs = 3.7313432835820895522388059701493
Code:
No of HLs   Activity	%
0	    131		100
1           65.5	50	
2           32.75	25
3           16.375	12.5
4           8.1875	6.25

So its somewhere between 12.5% and 6.25%... Cant quite remember how to do it exactly.

My guess would be from 3 > 4 the loss is 8.1875
So from 3 > 3.7313432835820895522388059701493 the lost would be 8.1875 * 0.7313432835820895522388059701493 = 5.9878731343283582089552238805949
So then take that away from the activity @ 3 HLs = 16.375 - 5.9878731343283582089552238805949 = 10.387126865671641791044776119406
The work out the percentage, so 10.387126865671641791044776119406 / 131 * 100 = 7.9291044776119402985074626865695 = 7.93% (2dp)
 
:smile: "Taking the number" is not a mathematical operation! You don't need to do anything with the "mass number" since you are only asked about percentage of the original amount, not "what mass was left".
Take the orginal amount to be 100%, multiply that by
(\frac{1}{2})^{\frac{30}{8.04}}
 
thanks to all who helped, upon thinking back and reviewing the formulas for this chapter i found the answer. here is the solution for anyone who cared...
 

Attachments

  • 3564564.jpg
    3564564.jpg
    14.4 KB · Views: 438
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top