A radioactive isotope with half life of 10 hours would decay to one fourth of its original amount in how many hours?
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Answer:
40 hrs
Explanation:
For a radioactive element at t =0, N = N0
and at t= T, N=(No/2) (Half –life)
From radioactive decay equation,
N = (No) e- k t where k = decay constant.
(No/2) = (No) e- k t
Taking log on both sides and cancelling No from both sides,
T (1/2) = (log (2))/ (k)
k = (log 2)/ (10)
Now, using this value of k, we get the time at which the activity of the element becomes (1/4th) of original activity, i.e.
N = (No/4)
Therefore t= ((1)/ (log2 /10)) x log ((No)/No/4))
= (10 hr/log 2) x log (24)
= (10 hr/log 2) x 4 log 2
T= 10 x 4
T = 40 hours
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