# 31.5 Half-life and activity  (Page 6/16)

 Page 6 / 16
$R={R}_{0}{e}^{-\mathrm{\lambda t}}\text{,}$

where ${R}_{0}$ is the activity at $t=0$ . This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation $R={R}_{0}{e}^{-\mathrm{\lambda t}}$ must be used to find $R$ .

## Phet explorations: alpha decay

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.

## Section summary

• Half-life ${t}_{1/2}$ is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei $N$ as a function of time is
$N={N}_{0}{e}^{-\mathrm{\lambda t}},$
where ${N}_{0}$ is the number present at $t=0$ , and $\lambda$ is the decay constant, related to the half-life by
$\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}.$
• One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity $R$ :
$R=\frac{\text{Δ}N}{\text{Δ}t}.$
• The SI unit for $R$ is the becquerel (Bq), defined by
$\text{1 Bq}=\text{1 decay/s.}$
• $R$ is also expressed in terms of curies (Ci), where
$1\phantom{\rule{0.25em}{0ex}}\text{Ci}=3\text{.}\text{70}×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq.}$
• The activity $R$ of a source is related to $N$ and ${t}_{1/2}$ by
$R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}.$
• Since $N$ has an exponential behavior as in the equation $N={N}_{0}{e}^{-\mathrm{\lambda t}}$ , the activity also has an exponential behavior, given by
$R={R}_{0}{e}^{-\mathrm{\lambda t}},$
where ${R}_{0}$ is the activity at $t=0$ .

## Conceptual questions

In a $3×{\text{10}}^{9}$ -year-old rock that originally contained some ${}^{\text{238}}\text{U}$ , which has a half-life of $4.5×{\text{10}}^{9}$ years, we expect to find some ${}^{\text{238}}\text{U}$ remaining in it. Why are ${}^{\text{226}}\text{Ra}$ , ${}^{\text{222}}\text{Rn}$ , and ${}^{\text{210}}\text{Po}$ also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?

Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.

Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?

Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?

Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)

To obtain the most precise value of BE from the equation $\text{BE=}\left[\text{ZM}\left({}^{1}\text{H}\right)+{\text{Nm}}_{n}\right]{c}^{2}-m\left({}^{A}X\right){c}^{2}$ , we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?

How does the finite range of the nuclear force relate to the fact that $\text{BE}/A$ is greatest for $A$ near 60?

## Problems&Exercises

Data from the appendices and the periodic table may be needed for these problems.

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of ${}^{\text{14}}\text{C}$ . Estimate the minimum age of the charcoal, noting that ${2}^{\text{10}}=\text{1024}$ .

57,300 y

#### Questions & Answers

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