A radioactive isotope decays according to the equation where Nt is the amount of
ID: 817103 • Letter: A
Question
A radioactive isotope decays according to the equation
where Nt is the amount of sample at time t, N0 is the original amount of sample, and ? is the decay constant. An unknown amount of an unidentified isotope is left to decay, and periodic measurements of the remaining isotope are recorded in the table below.
Time(days) amount of isotopes (Moles)
12.0 6.6
36.0 0.80
50.0 0.23
85.0 0.011
112.0 0.0010
1. What is ?, the decay constant?
2.What is N0, the original amount of radioactive material?
Explanation / Answer
We use the equation
?=ln(No/Nt)/t
where ?=Decay constant, t is time interval
No is initial concentration and Nt is concentration at given time interval
The equation can be re arranged as under
?=ln(No/Nt)/t
or
? * t =ln(No/Nt)
? * t =ln(No) - ln(Nt)
we are given values at various time intervals
we take the first set of values
t = 12 and Nt = 6.6
and apply the same to the equation above
? * 12 =ln(No) - ln(6.6)
? * 12 =ln(No) - 1.887
? * 12 + 1.887 =ln(No) ----- (1)
Now we take the second set of values
t = 36 and Nt = 0.8
and apply the same to the equation above
? * 36 =ln(No) - ln(0.8)
? * 36 =ln(No) - (- 0.223) = ln(No) + 0.223 -----(2)
now we take value of ln(No) from eq (1) above
and apply in eq (2)
we get
? * 36 = ? * 12 + 1.887 + 0.223
? * 36 - ? * 12 = 1.887 + 0.223
? * 24 = 2.11
? = 2.11/24
? = 0.0879 ( Answer) Decay Constant
we use value of ? = 0.0879 and apply in eq (1)
to solve for value of No
0.0879 * 12 + 1.887 =ln(No)
1.0548 + 1.887= ln(No)
Or
ln(No)=2.9418
No = 18.949 moles (Answer) the original amount of radioactive material