QuestionMay 26, 2025

19. After 32 days 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element? 40 grams 32 days 4 days 8 days

19. After 32 days 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element? 40 grams 32 days 4 days 8 days
19. After 32 days 5 milligrams of an 80-milligram sample of a radioactive isotope
remains unchanged. What is the half-life of this element?
40 grams
32 days
4 days
8 days

Solution
3.5(280 votes)

Answer

8 days Explanation 1. Use the decay formula The decay formula is N(t) = N_0 \cdot (1/2)^{t/T}, where N(t) is the remaining amount, N_0 is the initial amount, t is time, and T is the half-life. 2. Substitute known values Given N(t) = 5 mg, N_0 = 80 mg, and t = 32 days. Substitute these into the formula: 5 = 80 \cdot (1/2)^{32/T}. 3. Solve for T Divide both sides by 80: \frac{5}{80} = (1/2)^{32/T}. Simplify to 1/16 = (1/2)^{32/T}. Take the logarithm of both sides: \log(1/16) = \frac{32}{T} \cdot \log(1/2). Solve for T: T = \frac{32 \cdot \log(1/2)}{\log(1/16)}. 4. Calculate T Since \log(1/16) = 4 \cdot \log(1/2), then T = \frac{32}{4} = 8 days.

Explanation

1. Use the decay formula<br /> The decay formula is $N(t) = N_0 \cdot (1/2)^{t/T}$, where $N(t)$ is the remaining amount, $N_0$ is the initial amount, $t$ is time, and $T$ is the half-life.<br /><br />2. Substitute known values<br /> Given $N(t) = 5$ mg, $N_0 = 80$ mg, and $t = 32$ days. Substitute these into the formula: $5 = 80 \cdot (1/2)^{32/T}$.<br /><br />3. Solve for T<br /> Divide both sides by 80: $\frac{5}{80} = (1/2)^{32/T}$. Simplify to $1/16 = (1/2)^{32/T}$.<br /> Take the logarithm of both sides: $\log(1/16) = \frac{32}{T} \cdot \log(1/2)$.<br /> Solve for $T$: $T = \frac{32 \cdot \log(1/2)}{\log(1/16)}$.<br /><br />4. Calculate T<br /> Since $\log(1/16) = 4 \cdot \log(1/2)$, then $T = \frac{32}{4} = 8$ days.
Click to rate:

Similar Questions