Answer :
We start with the information that the mass of the sample was initially
[tex]$$m_i = 104.8\ \text{kg}$$[/tex]
at 12:02 P.M. and later measured as
[tex]$$m_f = 13.1\ \text{kg}$$[/tex]
at 4:11 P.M. for the same day.
Since radioactive decay follows the law
[tex]$$ m_f = m_i \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}, $$[/tex]
where [tex]\( T_{1/2} \)[/tex] is the half-life, we can determine the number of half-lives that occurred by looking at the ratio:
[tex]$$ \frac{m_f}{m_i} = \frac{13.1}{104.8}. $$[/tex]
Calculating the ratio gives approximately
[tex]$$ \frac{13.1}{104.8} \approx 0.125. $$[/tex]
Notice that
[tex]$$ 0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3. $$[/tex]
This means that during the period from 12:02 P.M. to 4:11 P.M., three half-life periods have passed, because
[tex]$$ \left(\frac{1}{2}\right)^3 = \frac{1}{8}. $$[/tex]
Next, we determine the time interval between the two measurements. First, convert each time into minutes after midnight.
For the initial time (12:02 P.M.):
[tex]$$ 12 \times 60 + 2 = 722\ \text{minutes}. $$[/tex]
For the later time (4:11 P.M. is equivalent to 16:11 in 24-hour time):
[tex]$$ 16 \times 60 + 11 = 971\ \text{minutes}. $$[/tex]
The difference in time between these two measurements is:
[tex]$$ 971 - 722 = 249\ \text{minutes}. $$[/tex]
Since three half-life periods have elapsed during these 249 minutes, the half-life [tex]\( T_{1/2} \)[/tex] is given by:
[tex]$$ T_{1/2} = \frac{249\ \text{minutes}}{3} = 83\ \text{minutes}. $$[/tex]
Finally, among the options provided, the only radioisotope with a half-life close to 83 minutes is Barium-139.
Therefore, the unknown radioisotope in the sample is Barium-139.
[tex]$$m_i = 104.8\ \text{kg}$$[/tex]
at 12:02 P.M. and later measured as
[tex]$$m_f = 13.1\ \text{kg}$$[/tex]
at 4:11 P.M. for the same day.
Since radioactive decay follows the law
[tex]$$ m_f = m_i \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}, $$[/tex]
where [tex]\( T_{1/2} \)[/tex] is the half-life, we can determine the number of half-lives that occurred by looking at the ratio:
[tex]$$ \frac{m_f}{m_i} = \frac{13.1}{104.8}. $$[/tex]
Calculating the ratio gives approximately
[tex]$$ \frac{13.1}{104.8} \approx 0.125. $$[/tex]
Notice that
[tex]$$ 0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3. $$[/tex]
This means that during the period from 12:02 P.M. to 4:11 P.M., three half-life periods have passed, because
[tex]$$ \left(\frac{1}{2}\right)^3 = \frac{1}{8}. $$[/tex]
Next, we determine the time interval between the two measurements. First, convert each time into minutes after midnight.
For the initial time (12:02 P.M.):
[tex]$$ 12 \times 60 + 2 = 722\ \text{minutes}. $$[/tex]
For the later time (4:11 P.M. is equivalent to 16:11 in 24-hour time):
[tex]$$ 16 \times 60 + 11 = 971\ \text{minutes}. $$[/tex]
The difference in time between these two measurements is:
[tex]$$ 971 - 722 = 249\ \text{minutes}. $$[/tex]
Since three half-life periods have elapsed during these 249 minutes, the half-life [tex]\( T_{1/2} \)[/tex] is given by:
[tex]$$ T_{1/2} = \frac{249\ \text{minutes}}{3} = 83\ \text{minutes}. $$[/tex]
Finally, among the options provided, the only radioisotope with a half-life close to 83 minutes is Barium-139.
Therefore, the unknown radioisotope in the sample is Barium-139.
