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Nuclear Physics. Help Me Solve This Problem.?
A radioactive material contains two radionuclides, one has a half-life of 1 day and another has a half-life of 8 days. Initially the radioactivity of the shortlived nuclide is 128 times greater than that of the longlived nuclide. When will their activities be equal?
I know the answer :8 Days. I have the solution as well but it's too long. I would like to have the shortest, easiest and simplest solution to this problem. Thanx.
1 Answer
- Old Science GuyLv 71 decade agoFavorite Answer
we may assume that the activity is proportional to the amount of substance present (N)
the most common half-life equation is
N = No * e^(-0.693t/k)
N= amount present at time = t
No - original amount at t=0
k = half-life in compatible time units (days in this problem)
so simply set the two amounts (N) equal and solve for t
128No * e^(-0.693t/1) = 1No * e^(-0.693t/8)
cancel No
128 * e^(-0.693t/1) = 1 * e^(-0.693t/8)
take the natural log
ln128 + (-0.693t/1) = (-0.693t/8)
ln128 + 8(-0.693t/8) = (-0.693t/8) (common denominators)
7(0.693t/8) = ln128 = 4.852 (rearranged - note the change of sign as we changed sides)
t= 8.00 day
there are other equations you may choose
N = No log(-0.301t/k)
or
N = No (1/2)^(t/k)
