The radioactive nuclide undergoes first-order decay with a half-life of 9.31 min.If a quantity of Is produced,what fraction remains after 82.5 seconds?

Question

Quizplus · Accepted Answer

We can use the equation for first-order decay:

N(t) = N0*e^(-kt)

where N0 is the initial quantity, N(t) is the quantity at time t, k is the decay constant (k= ln2/t1/2) and t is the time.

We are given the half-life (t1/2) of the nuclide as 9.31 min. Therefore, the decay constant is:

k = ln2/t1/2 = ln2/9.31 = 0.07449 min^-1

We can use this value of k to find the quantity remaining after 82.5 seconds (1.375 min):

N(1.375) = N0*e^(-0.07449*1.375)

N(1.375) = 0.0973 * N0

Therefore, the fraction of the nuclide remaining after 82.5 seconds is 0.0973 or 9.73%. The closest answer choice is (E) 0.0973.

The Radioactive Nuclide Undergoes First-Order Decay with a Half-Life