please use the values in the resources listed below instead of the textbook values. the isotope…

please use the values in the resources listed below instead of the textbook values. the isotope $_{38}^{90}sr$ is one of the hazardous species in the residues from nuclear power generation. the strontium in a 0.419 g sample diminishes to 0.279 g in 17.0 y. calculate the half - life (in y).

please use the values in the resources listed below instead of the textbook values. the isotope $_{38}^{90}sr$ is one of the hazardous species in the residues from nuclear power generation. the strontium in a 0.419 g sample diminishes to 0.279 g in 17.0 y. calculate the half - life (in y).

Answer

Answer:

(31.5)

Explanation:

Step1: Use the radioactive decay formula

The radioactive decay formula is (N = N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}), where (N) is the final amount, (N_0) is the initial amount, (t) is the time elapsed, and (T_{1/2}) is the half - life. Given (N_0=0.419\space g), (N = 0.279\space g), and (t = 17.0\space y). Substitute these values into the formula: (0.279=0.419\left(\frac{1}{2}\right)^{\frac{17.0}{T_{1/2}}}).

Step2: Rearrange the formula

First, divide both sides by (0.419): (\frac{0.279}{0.419}=\left(\frac{1}{2}\right)^{\frac{17.0}{T_{1/2}}}). (0.666=\left(\frac{1}{2}\right)^{\frac{17.0}{T_{1/2}}}). Take the natural logarithm of both sides: (\ln(0.666)=\ln\left[\left(\frac{1}{2}\right)^{\frac{17.0}{T_{1/2}}}\right]). Using the property of logarithms (\ln(a^b)=b\ln(a)), we get (\ln(0.666)=\frac{17.0}{T_{1/2}}\ln\left(\frac{1}{2}\right)). Since (\ln\left(\frac{1}{2}\right)=-\ln(2)\approx - 0.693) and (\ln(0.666)\approx-0.404).

Step3: Solve for (T_{1/2})

We have (-0.404=\frac{17.0}{T_{1/2}}(- 0.693)). Cross - multiply: (-0.404T_{1/2}=-0.693\times17.0). (-0.404T_{1/2}=-11.781). Then (T_{1/2}=\frac{11.781}{0.404}\approx31.5\space y).