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Expert-verified Found in: Page 11 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A generation is about one-third of a lifetime. Approximately how many generations have passed since the year $${\bf{0}}$$ AD?

The required number of generations have passed since the year $$0$$ AD is $$95$$.

See the step by step solution

## Step 1: Defining sample half-life

The time taken by the sample to decay to half the original number of the nuclei is called the half-life of the sample.

The expression of the half-life of a radioactive isotope is given as follows.

$${t_{{\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\\} \!\lower0.7ex\hbox{2}}}} = \frac{{0.693}}{\lambda }$$

Here, $$\lambda$$ is the activity of the sample.

The half-life of the time is in the order as,

$${T_{{\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\\} \!\lower0.7ex\hbox{2}}}} = {10^9}{\rm{ }}s$$

The half-life of the a generation is twice the half-life.

\begin{aligned}{c}1{\rm{ }}life - time &= 2\left( {{T_{{1 \mathord{\left/ {\vphantom {1 2}} \right.\\} 2}}}} \right)\\ &= 2\left( {{{10}^9}{\rm{ }}s} \right)\end{aligned}

## Step 2: Calculating generations passed in a given year

The time elapsed since $$0{\rm{ }}AD$$ is $$2019$$ years. The expression for the number of generations passed is,

$$Number{\rm{ }}of{\rm{ generations}} = \frac{{2019}}{{\frac{1}{3} \times life{\rm{ }}time}}$$

Substitute $$2\left( {{{10}^9}{\rm{ }}s} \right)$$ for $$1{\rm{ }}life - time$$ in the above equation.

\begin{aligned}{c}Number{\rm{ }}of{\rm{ generations}} &= \frac{{3 \times \left( {2019{\rm{ years}}} \right)\left( {\frac{{3.15 \times {{10}^9}{\rm{ }}s}}{{1{\rm{ }}year}}} \right)}}{{2\left( {{{10}^9}{\rm{ }}s} \right)}}\\ &= 95.39\\ \approx 95\end{aligned}

## Step 3: Deriving conclusions

Hence, the required number of generations that have passed since the year $$0$$ AD is $$95$$. ### Want to see more solutions like these? 