A sample of a radioisotope shows an activity of 999 disintegrations per minute due to beta decay.If after 1.10 years the activity is 952 disintegrations per minute, what is the half-life of this radioisotope?

Question

Quizplus · Accepted Answer

We can use the formula for radioactive decay:

A = A0 * (1/2)^(t/T)

where A is the activity at time t, A0 is the initial activity, T is the half-life, and t is the time elapsed.

We know that A0 = 999, A = 952, and t = 1.10 years. Substituting these values, we get:

952 = 999 * (1/2)^(1.10/T)

Dividing both sides by 999, we get:

(952/999) = (1/2)^(1.10/T)

Taking the logarithm of both sides, we get:

log(952/999) = (1.10/T) * log(1/2)

Solving for T, we get:

T = (1.10/log(2)) * (-1/log(952/999))

T ≈ 15.8 years

Therefore, the half-life of this radioisotope is 15.8 years, which is option D.

A Sample of a Radioisotope Shows an Activity of 999