Physics

Physics

Problem

A semilogarithmic plot of the natural logarithm of activity, ln⁡A\ln A, versus time tt is provided for isotope W. The plotted straight line intersects the following points exactly: • (t=0 h,  ln⁡A=9.21)(t = 0\,\text{h},\; \ln A = 9.21) • (t=4.0 h,  ln⁡A=7.83)(t = 4.0\,\text{h},\; \ln A = 7.83) • (t=8.0 h,  ln⁡A=6.45)(t = 8.0\,\text{h},\; \ln A = 6.45) Assume these points lie on the same line for all times measured. The initial activity corresponds to A0=10,000 counts s−1A_0 = 10{,}000\,\text{counts\,s}^{-1} because ln⁡10,000=9.21\ln 10{,}000 = 9.21. Respond to each of the following prompts. (a) Use the plotted data to obtain a numerical value for the decay constant λ\lambda of isotope W, including units. (b) Calculate the half-life of isotope W from your value of λ\lambda. (c) Determine the time at which the activity will first reach 250 counts s−1250\,\text{counts\,s}^{-1}. Show all reasoning steps. (d) Explain two advantages of plotting ln⁡A\ln A versus tt rather than AA versus tt when one wishes to determine decay parameters from experimental data.
Intuitive explanation

Radioactive decay follows an exponential law, and plotting ln⁡A\ln A versus tt transforms that exponential into a straight line—the slope of that line is directly the decay constant λ\lambda. Once you find λ\lambda from the graph's slope, you can calculate the half-life using the relationship t1/2=ln⁡2λt_{1/2} = \frac{\ln 2}{\lambda}, and predict activity at any future time using A(t)=A0e−λtA(t) = A_0 e^{-\lambda t}.

Intuitive explanation

Radioactive decay follows an exponential law, and plotting ln⁡A\ln A versus tt transforms that exponential into a straight line—the slope of that line is directly the decay constant λ\lambda. Once you find λ\lambda from the graph's slope, you can calculate the half-life using the relationship t1/2=ln⁡2λt_{1/2} = \frac{\ln 2}{\lambda}, and predict activity at any future time using A(t)=A0e−λtA(t) = A_0 e^{-\lambda t}.

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