Question

A radioactive isotope with half life of 10 hours would decay to one fourth of its original amount in how many hours?

​

Answer 1

Answer:

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