1) the radioactive element radium decays according to the
formula N=N0e^(-kt)
where N0 is the initial amount(in grams)at time t=0 and
k is a constant.The half-life of radium is 1590years.
a)find the value of the constant k cor. to 3sig.fig.
the ans. is 0.000436
pls show clear steps.thz
maths&stat. logarithmicfunctio
2010-04-12 2:16 am
回答 (2)
2010-04-12 2:25 am
✔ 最佳答案
當 t = 1590，它已衰變了一半,即 N = No/2
No/2 = Noe^(-1590k)
e^(-1590k) = 1/2
-1590 k = ln(1/2)
k = 0.000436
2010-04-12 2:28 am
For the meaning of a half-time, it is the time required for the radioactive element to decay to half of its initial amount.
So, at t = t1/2, N = 1/2 N0
So,
1/2 N0 = N0e^(-1590k)
1/2 = e^(-1590k)
ln(1/2) = -1590k
k = (1/1590)ln2
k = 0.000436 (cor. to 3 sig. fig.)
So, at t = t1/2, N = 1/2 N0
So,
1/2 N0 = N0e^(-1590k)
1/2 = e^(-1590k)
ln(1/2) = -1590k
k = (1/1590)ln2
k = 0.000436 (cor. to 3 sig. fig.)
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