Use the half-life information to complete each table. (The formula 𝒩=𝒩0 e^k t is not required.)

Name: Problem 24, Chapter 5: Precalculus
Uploaded: 2020-08-11T16:44:29-08:00
Duration: 1 min 5 s
Channel: Carson Merrill
Description: Use the half-life information to complete each table. (The formula 𝒩=𝒩0 e^k t is not required.) (a) Polonium-210: half-life =138.4 days t (years) 0 138.4 276.8 415.2 𝒩 (grams) 0.4 0.025 (b) Polonium-214: half-life =1.63 ×10^-4 second t (seconds) 0 6.52 × 10^-4 𝒩 (grams) 0.1 0.05 0.025 0.0125

step 1 So for Part A, we would have .4, then point two, then point one, then .05, and then in the top r

Question

Use the half-life information to complete each table. (The formula $\mathcal{N}=\mathcal{N}_{0} e^{k t}$ is not required.) (a) Polonium-210: half-life $=138.4$ days $$\begin{array}{lllll}\hline t \text { (years) } & 0 & 138.4 & 276.8 & 415.2 \\\mathcal{N} \text { (grams) } & 0.4 & & & 0.025 \\\hline\end{array}$$ (b) Polonium-214: half-life $=1.63 \times 10^{-4}$ second $$\begin{array}{lllll}\hline t \text { (seconds) } & 0 & & & 6.52 \times 10^{-4} \\\mathcal{N} \text { (grams) } & 0.1 & 0.05 & 0.025 & 0.0125 & \\\hline\end{array}$$

   Use the half-life information to complete each table. (The formula $\mathcal{N}=\mathcal{N}_{0} e^{k t}$ is not required.)
(a) Polonium-210: half-life $=138.4$ days
$$\begin{array}{lllll}\hline t \text { (years) } & 0 & 138.4 & 276.8 & 415.2 \\\mathcal{N} \text { (grams) } & 0.4 & & & 0.025 \\\hline\end{array}$$
(b) Polonium-214: half-life $=1.63 \times 10^{-4}$ second
$$\begin{array}{lllll}\hline t \text { (seconds) } & 0 & & & 6.52 \times 10^{-4} \\\mathcal{N} \text { (grams) } & 0.1 & 0.05 & 0.025 & 0.0125 & \\\hline\end{array}$$

Precalculus

David Cohen,… 4th Edition

Chapter 5, Problem 24

Instant Answer

Solved by Expert Carson Merrill

08/11/2020

Step 1

4 days. This means that every 138.4 days, the amount of Polonium-210 will be halved. Show more…

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Use the half-life information to complete each table. (The formula $\mathcal{N}=\mathcal{N}_{0} e^{k t}$ is not required.) (a) Polonium-210: half-life $=138.4$ days $$\begin{array}{lllll}\hline t \text { (years) } & 0 & 138.4 & 276.8 & 415.2 \\\mathcal{N} \text { (grams) } & 0.4 & & & 0.025 \\\hline\end{array}$$ (b) Polonium-214: half-life $=1.63 \times 10^{-4}$ second $$\begin{array}{lllll}\hline t \text { (seconds) } & 0 & & & 6.52 \times 10^{-4} \\\mathcal{N} \text { (grams) } & 0.1 & 0.05 & 0.025 & 0.0125 & \\\hline\end{array}$$

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Key Concepts

Half-Life

This is the time required for a given quantity of a radioactive substance to decay to half its initial amount. It is a key measure in understanding the rate of decay in nuclear processes and is used to calculate the remaining quantity of a substance after multiple decay cycles.

Exponential Decay

This concept describes a process where the decrease of a quantity is proportional to its current value, resulting in a rapid decline that slows over time. In the context of radioactive decay, it explains how, after each half-life, the quantity reduces by a factor of one half.

Decay Constant

This parameter quantifies the probability per unit time that a nucleus will decay, and it is mathematically related to the half-life. Although sometimes the calculation does not require the use of the decay constant, understanding its connection to half-life and exponential decay is fundamental in the study of nuclear decay processes.

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