Question

A freshly prepared sample of a radioisotope of half life 1386s has activity 103 disintegrations per second. Given that ln2 = 0.693 , the fraction of the initial number of nuclei ( expressed in nearest integer percentage) that will decay in the first 80s after preparation of the sample is

Answer 1

Answer:

The answer will be 4%

Explanation:

According to the problem the half life of the sample is given as well the it is also given that its activity is 10^3 disintegrations per second.

Now the formula is

N = N0 e^−λt [ where n is the number of nuclei undecayed , N0 = number of nuclei initially]

Now fraction = (N0 - N)/N0 [ where N0 - N is  decayed nuclei]

                         = 1 - N/N0 = 1 - N0 e^−λt/N0 = 1-  e^−λt

Now λ = 0.693/ t(1/2) = 0.693/1386

Therefore,

fraction = 1-  e^−0.693/1386  x 80 = 0.04(apprx)

Hence the fraction is 4%