An isotope has an initial activity of 120 Bq. 6 days later its activity is

Explanation

The equation that establishes a relationship between the amountleft undecayed,

A, the initial amount, A0, and thenumber 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