Answer:
(C) ln [Bi]
Explanation:
Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.
Answer:
Option C. ln[Bi]
Explanation:
The nuclear equation of first-order radioactive decay of Bi to Po is:
²¹⁴Bi₈₃ → ²¹⁴Po₈₄ + e⁻
The radioactive decay is expressed by the following equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex] (1)
where Nt: is the number of particles at time t, No: is the initial number of particles, and λ: is the decay constant.
To plot the variation of the quantities in function of time, we need to solve equation (1) for t:
[tex] Ln(\frac{N_{t}}{N_{0}}) = -\lambda t [/tex] (2)
Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:
[tex] [Bi] = \frac {N particles}{N_{A} * V} [/tex] (3)
[tex] [Bi]_{0} = \frac {N_{0} particles}{N_{A} * V} [/tex] (4)
where [tex]N_{A}[/tex]: si the Avogadro constant and V is the volume.
Now, introducing equations (3) and (4) into (2), we have:
[tex] Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t [/tex]
[tex] Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t [/tex] (5)
Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:
[tex] Ln ([Bi]) = -\lambda t + Ln([Bi]_{0}) [/tex]
Therefore, the correct answer is option C. ln[Bi].
I hope it helps you!
