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/t_{1/2} = ?? when t_{1/2}= 30 years.
ln(N_{o}/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(N_{o}/N)=kt where
N_{o} = 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 N_{o} 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 = t_{1/2} and solve for k.
ln(100/50)=k*t_{1/2}
ln(1/2)= k*t_{1/2}
0.693 = k*t_{1/2} and
k = 0.693/t_{1/2}.
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.
The isotope caesium-137, which has a half life of 30 years, is a product of neuclear 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.
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the isotope caesium-137 has a half life of 30 years how long will it take for the amount of this isotope in a sample of caesium to decay to 1/16th of its origional amount. thanks
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