Asked by Nancy
The half life of U238 is 4.5 x 10^9 yr. A sample of rock of 1.6 g produces 29 dis/sec. Assuming all radioactivity is due to U238 find percent by mass of U 238.
I have tried
ln(t)/ln(0)= -kt and still cannot get the answer
What I get is
A) 4.5x10^9 x365x60x60x24=1.419x10^17 for half life
B) .693/1.419x10^17=4.883x10-18
C) ln t/ln 29= -(4.883x10^-18)
That is the farthest I've gotten. Can someone help?
I have tried
ln(t)/ln(0)= -kt and still cannot get the answer
What I get is
A) 4.5x10^9 x365x60x60x24=1.419x10^17 for half life
B) .693/1.419x10^17=4.883x10-18
C) ln t/ln 29= -(4.883x10^-18)
That is the farthest I've gotten. Can someone help?
Answers
Answered by
Graham
Use Nt/N0 = e^{-kt}
1[yr] = a = 31557600[s]
(1/2) = e^(-kh)
Where:
Half life: h= 4.5e9 * a = 1.420092E+17[s]
k = -ln(0.5)/1.420092E+17
.: k = 4.8810019E-18[/s]
Okay, so far. To continue...
Assume there is n atoms of U238 in the sample. t=1 second latter there is n-29. (via 29 disintegrations per second)
(n-29)/n = e^{-kt}
1 - (29/n) = e^{-kt}
29/n = 1-e^{-kt}
n = 29/(1-e^{-kt})
n = 29/(1-exp(-4.8810019E-18))
n = 5.9414031E+18[atoms]
Next: find the weight of this number.
Given:
1[mole]=6.02214129(27)E+23[atoms]
U[238] weighs 238.05078822[g/mol]
Express as a percentage of the sample weight (1.6[g]).
1[yr] = a = 31557600[s]
(1/2) = e^(-kh)
Where:
Half life: h= 4.5e9 * a = 1.420092E+17[s]
k = -ln(0.5)/1.420092E+17
.: k = 4.8810019E-18[/s]
Okay, so far. To continue...
Assume there is n atoms of U238 in the sample. t=1 second latter there is n-29. (via 29 disintegrations per second)
(n-29)/n = e^{-kt}
1 - (29/n) = e^{-kt}
29/n = 1-e^{-kt}
n = 29/(1-e^{-kt})
n = 29/(1-exp(-4.8810019E-18))
n = 5.9414031E+18[atoms]
Next: find the weight of this number.
Given:
1[mole]=6.02214129(27)E+23[atoms]
U[238] weighs 238.05078822[g/mol]
Express as a percentage of the sample weight (1.6[g]).
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.