Question

this chart show the amount of radioactivity measured from three unknown isotopes.

isotope	starting weight	ending measured weight	half - life
b	20	2.5	2 days
c	45	22.5	10 days

which statement describes each of the isotopes?
isotope a was measured at day 24, isotope b was measured at day 6, and isotope c was measured at day 10.
isotope a was measured at day 18, isotope b was measured at day 4, and isotope c was measured at day 5.
isotope a was measured at day 24, isotope b was measured at day 8, and isotope c was measured at day 20.
isotope a was measured at day 30, isotope b was measured at day 6, and isotope c was measured at day 10.

Explanation:

Step1: Recall radioactive - decay formula

The formula for radioactive decay is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the ending amount, $N_0$ is the starting amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. We can re - arrange it to solve for $t$: $t=T_{1/2}\times\frac{\log(\frac{N}{N_0})}{\log(\frac{1}{2})}$.

Step2: Calculate time for Isotope A

For Isotope A, $N_0 = 95$, $N = 5.9$, and $T_{1/2}=6$ days.
$t_A=6\times\frac{\log(\frac{5.9}{95})}{\log(\frac{1}{2})}=6\times\frac{\log(0.0621)}{\log(0.5)}=6\times\frac{- 1.207}{-0.301}=24$ days.

Step3: Calculate time for Isotope B

For Isotope B, $N_0 = 20$, $N = 2.5$, and $T_{1/2}=2$ days.
$t_B=2\times\frac{\log(\frac{2.5}{20})}{\log(\frac{1}{2})}=2\times\frac{\log(0.125)}{\log(0.5)}=2\times\frac{-0.903}{-0.301}=6$ days.

Step4: Calculate time for Isotope C

For Isotope C, $N_0 = 45$, $N = 22.5$, and $T_{1/2}=10$ days.
$t_C=10\times\frac{\log(\frac{22.5}{45})}{\log(\frac{1}{2})}=10\times\frac{\log(0.5)}{\log(0.5)} = 10$ days.

Answer:

Isotope A was measured at day 24, Isotope B was measured at day 6, and Isotope C was measured at day 10.