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
The correct answer is: B
Explanation
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:
\(\frac{1}{2^n} * A_0 = A\)
\(\frac{1}{2^n} * 120 = 15\)
\(\frac{1}{2^n}\) = \(\frac{15}{120}\)
\(\frac{1}{2^n}\) = \(\frac{1}{8}\)
\(\frac{1}{2^n}\) = \(\frac{1}{2^3}\)
\(2^n = 2^3\)
n = 3
:6days ÷ 3 = 2days
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