# science

The isotope caesium-137, which has a half-life of 30 years, is a product of nuclear power plants. How long will it take for the amount of this isotope in a sample of caesium to decay to one-sixteenth of its original amount?

k = 0.693/t1/2 = ?? when t1/2= 30 years.

ln(No/N) = kt
ln(1/16)= kt. Plug in the value from k, from above, and solve for t.

how did you find 0.693? and is this equation then...

0.693/30 = 0.231 = k (when t=30)

and [ln(1/16)]/0.231 = t = -12.0025...

I AM LOST, PLS HELP

You're doing fine except you punched in the wrong number somewhere. 0.693/30 = 0.0231 and not 0.231.
Then ln(1/16)= 0.0231*t
1.06=0.0231*t
t=1.06/0.0231 = 46.1 years for the material to decay to 1/16 its original value.

As for the 0.693 number, I have it memorized; however, you can calculate it if you wish. It comes from the original equation of
ln(No/N)=kt where
No = the initial number of atoms of a material at time zero.
N= the number at the present/future time.
k is a constant and t is the time.
What we do, to solve for the constant, k, is to call No say 100, then when exactly half has decayed we will have 50 left (and the 50 will be N), and since half has decayed then t = t1/2 and solve for k.
ln(100/50)=k*t1/2
ln(1/2)= k*t1/2
0.693 = k*t1/2 and
k = 0.693/t1/2.
In this manner, then, given a half life, as in your poblem, of 30 years, we can calculate k, then use the k in the ln part of the equation and solve for whatever unknown is appropriate. Here it was t that was the unknown.
I hope this helps.

Makes a heap more sense, thank-you.

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