Decay of [latex]{ }^{18} F[/latex] For [latex]{ }^{18} F[/latex] F of the preceding example , how much time elapses before only 1% of the original sample remains? ORGANIZE AND PLAN The fraction of nuclei remaining is related to time by the radioactive decay law Equation 25.5 describes the radioactive decay: [latex]\left.N=N_{0} e^{-\lambda t}\right),[/latex] whereas in the preceding example [latex]\lambda=\ln 2 / t_{1 / 2} [/latex]. [latex]\text { We want } 1 \% \text { of the original sample to remain, so } N=N_{0} / 100 [/latex]. [latex]N=N_{0} e^{-\lambda t}[/latex] (Radioactive decay) (25.5). [latex]\text { Known: } t_{1 / 2}=110 min =6600 s ; N / N_{0}=0.010 [/latex].

[latex]\text { With } 1 \% \text { of the nuclei remaining, } N_{0} / 100=N_{0} e^{-\lambda t} \text { or, using }[/latex] [latex]e^{-x}=1 / e^{x}, e^{\lambda t}=100[/latex] To solve, take the natural logarithm of both sides: [latex]\ln \left(e^{\lambda t}\right)=\lambda t=\ln 100 \text {. Solving for } t[/latex]. [latex]t=\frac{\ln 100}{\lambda}=\frac{\ln 100}{\ln 2} t_{1 / 2}=\frac{\ln 100}{\ln 2}(6600 s )=4.38 \times 10^{4} s[/latex]. REFLECT That's just about 12 h another indication that this short-half- life isotope doesn't stay around very long.

Question 25.7: Decay of ^18F For ^18F of the preceding example, how much ti...

Essential College Physics

Decay of ${ }^{18} F$

For ${ }^{18} F$ F of the preceding example, how much time elapses before only 1% of the original sample remains?

ORGANIZE AND PLAN The fraction of nuclei remaining is related to time by the radioactive decay law Equation 25.5 describes the radioactive decay: $\left.N=N_{0} e^{-\lambda t}\right),$ whereas in the preceding example $\lambda=\ln 2 / t_{1 / 2}$ .

$\text { We want } 1 \% \text { of the original sample to remain, so } N=N_{0} / 100$ .

$N=N_{0} e^{-\lambda t}$ (Radioactive decay) (25.5).

$\text { Known: } t_{1 / 2}=110 min =6600 s ; N / N_{0}=0.010$ .

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$\text { With } 1 \% \text { of the nuclei remaining, } N_{0} / 100=N_{0} e^{-\lambda t} \text { or, using }$ $e^{-x}=1 / e^{x}, e^{\lambda t}=100$ To solve, take the natural logarithm of both sides: $\ln \left(e^{\lambda t}\right)=\lambda t=\ln 100 \text {. Solving for } t$ .

$t=\frac{\ln 100}{\lambda}=\frac{\ln 100}{\ln 2} t_{1 / 2}=\frac{\ln 100}{\ln 2}(6600 s )=4.38 \times 10^{4} s$ .

REFLECT That’s just about 12 h another indication that this short-half- life isotope doesn’t stay around very long.