An isotope has an initial activity of 120 Bq. 6 days late...
An isotope has an initial activity of 120 Bq. 6 days later its activity is 15 Bq. The half-life is?
3 days
2 days
1 day
4 days
Correct answer is B
The equation that establishes a relationship between the amount left undecayed,
A, the initial amount, A0, and the number of half-lives that pass in a period of time t looks like this:
12n∗A0=A
12n∗120=15
12n = 15120
12n = 18
12n = 123
2n=23
n = 3
:6days ÷ 3 = 2days