Question

214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle, as represented by the nuclear equation above. Which of the following quantities plotted versus time will produce a straight line?(A) [Bi](B) [Po](C) ln[Bi](D) 1/[Bi]

Answer 1

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:

$N_{t} = N_{0}e^{-\lambda t}$      (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:

$Ln(\frac{N_{t}}{N_{0}}) = -\lambda t$      (2)

Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:

$[Bi] = \frac {N particles}{N_{A} * V}$                            (3)  

$[Bi]_{0} = \frac {N_{0} particles}{N_{A} * V}$               (4)  

where $N_{A}$: si the Avogadro constant and V is the volume.

Now, introducing equations (3) and (4) into (2), we have:

$Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t$  

$Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t$      (5)

Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:

$Ln ([Bi]) = -\lambda t + Ln([Bi]_{0})$      

Therefore, the correct answer is option C. ln[Bi].

I hope it helps you!          

Answer 2

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.