An isotope has an initial activity of 120 Bq. 6 days later its activity is 15 Bq. The half-life is?

Answer Details

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 $\frac{\mathrm{<br>}}{}<br>$

12n∗120=15 $<br>$

12n $\frac{\mathrm{<br>}}{}$ = 15120 $\frac{\mathrm{<br>}}{}$

12n $\frac{\mathrm{<br>}}{}$ = 18 $\frac{\mathrm{<br>}}{}$

12n = 123 $\frac{\mathrm{<br>}}{}$

2n=23 ${\mathrm{<br>}}^{}<br>$

n = 3
:6days ÷ 3 = 2days