Answer :
We begin by assuming that the mass of the radioisotope decays exponentially according to
[tex]$$
M(t) = M_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}},
$$[/tex]
where
- [tex]$M_0$[/tex] is the initial mass,
- [tex]$M(t)$[/tex] is the mass at time [tex]$t$[/tex], and
- [tex]$t_{1/2}$[/tex] is the half-life.
Rachel measured the mass at two different times:
- At 12:02 P.M., the mass was [tex]$104.8\text{ kg}$[/tex].
- At 4:11 P.M., the mass was [tex]$13.1\text{ kg}$[/tex].
Step 1. Calculate the elapsed time
Convert each time into minutes:
- 12:02 P.M. corresponds to [tex]$12 \times 60 + 2 = 722$[/tex] minutes.
- 4:11 P.M. (which is 16:11 in 24-hour time) corresponds to [tex]$16 \times 60 + 11 = 971$[/tex] minutes.
The elapsed time is
[tex]$$
t = 971 - 722 = 249 \text{ minutes}.
$$[/tex]
It can also be expressed in hours:
[tex]$$
t = \frac{249}{60} \approx 4.15 \text{ hours}.
$$[/tex]
Step 2. Determine the decay ratio
The ratio of final mass to initial mass is
[tex]$$
\frac{M(t)}{M_0} = \frac{13.1}{104.8}.
$$[/tex]
Notice that
[tex]$$
\frac{104.8}{8} = 13.1,
$$[/tex]
so the ratio simplifies to
[tex]$$
\frac{M(t)}{M_0} = \frac{1}{8}.
$$[/tex]
Since [tex]$8 = 2^3$[/tex], we can write
[tex]$$
\frac{1}{8} = \left(\frac{1}{2}\right)^3.
$$[/tex]
Step 3. Relate the ratio to the half-life
From the decay law, we have
[tex]$$
\frac{M(t)}{M_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}.
$$[/tex]
Thus,
[tex]$$
\left(\frac{1}{2}\right)^{\frac{249}{t_{1/2}}} = \left(\frac{1}{2}\right)^3.
$$[/tex]
Since the bases are the same, the exponents must be equal:
[tex]$$
\frac{249}{t_{1/2}} = 3.
$$[/tex]
Solving for [tex]$t_{1/2}$[/tex], the half-life, we obtain
[tex]$$
t_{1/2} = \frac{249}{3} = 83 \text{ minutes}.
$$[/tex]
Expressed in hours, this is approximately
[tex]$$
t_{1/2} \approx \frac{83}{60} \approx 1.38 \text{ hours}.
$$[/tex]
Step 4. Identify the radioisotope
Among the given options, the radioisotope with a half-life of about [tex]$83$[/tex] minutes is [tex]$\boxed{\text{Barium-139}}$[/tex].
[tex]$$
M(t) = M_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}},
$$[/tex]
where
- [tex]$M_0$[/tex] is the initial mass,
- [tex]$M(t)$[/tex] is the mass at time [tex]$t$[/tex], and
- [tex]$t_{1/2}$[/tex] is the half-life.
Rachel measured the mass at two different times:
- At 12:02 P.M., the mass was [tex]$104.8\text{ kg}$[/tex].
- At 4:11 P.M., the mass was [tex]$13.1\text{ kg}$[/tex].
Step 1. Calculate the elapsed time
Convert each time into minutes:
- 12:02 P.M. corresponds to [tex]$12 \times 60 + 2 = 722$[/tex] minutes.
- 4:11 P.M. (which is 16:11 in 24-hour time) corresponds to [tex]$16 \times 60 + 11 = 971$[/tex] minutes.
The elapsed time is
[tex]$$
t = 971 - 722 = 249 \text{ minutes}.
$$[/tex]
It can also be expressed in hours:
[tex]$$
t = \frac{249}{60} \approx 4.15 \text{ hours}.
$$[/tex]
Step 2. Determine the decay ratio
The ratio of final mass to initial mass is
[tex]$$
\frac{M(t)}{M_0} = \frac{13.1}{104.8}.
$$[/tex]
Notice that
[tex]$$
\frac{104.8}{8} = 13.1,
$$[/tex]
so the ratio simplifies to
[tex]$$
\frac{M(t)}{M_0} = \frac{1}{8}.
$$[/tex]
Since [tex]$8 = 2^3$[/tex], we can write
[tex]$$
\frac{1}{8} = \left(\frac{1}{2}\right)^3.
$$[/tex]
Step 3. Relate the ratio to the half-life
From the decay law, we have
[tex]$$
\frac{M(t)}{M_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}.
$$[/tex]
Thus,
[tex]$$
\left(\frac{1}{2}\right)^{\frac{249}{t_{1/2}}} = \left(\frac{1}{2}\right)^3.
$$[/tex]
Since the bases are the same, the exponents must be equal:
[tex]$$
\frac{249}{t_{1/2}} = 3.
$$[/tex]
Solving for [tex]$t_{1/2}$[/tex], the half-life, we obtain
[tex]$$
t_{1/2} = \frac{249}{3} = 83 \text{ minutes}.
$$[/tex]
Expressed in hours, this is approximately
[tex]$$
t_{1/2} \approx \frac{83}{60} \approx 1.38 \text{ hours}.
$$[/tex]
Step 4. Identify the radioisotope
Among the given options, the radioisotope with a half-life of about [tex]$83$[/tex] minutes is [tex]$\boxed{\text{Barium-139}}$[/tex].
