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Expert-verified Found in: Page 33 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Question: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of$${\bf{1}}{{\bf{0}}^{ - {\bf{22}}}}{\rm{ }}{\bf{s}}$$.

Hence, the lifetime of an unstable atomic nucleus is $$4.0 \times {10^{ - 32}}$$ times longer.

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 average lifetime of an unstable atomic nucleus is in the order is

$${t_{nucleus}} = {10^{ - 22}}{\rm{ }}s$$

The average lifetime of human is being $$80{\rm{ }}years$$.

In seconds, a life time is

\begin{aligned}{}{t_{human}} = 80{\rm{ }}years \times \frac{{365.2{\rm{ }}days}}{{1{\rm{ }}years}} \times \frac{{24{\rm{ }}hr}}{{1{\rm{ }}day}} \times \frac{{{\rm{3600 sec}}}}{{1{\rm{ hr}}}}\\ \approx 2.5 \times {10^9}{\rm{ }}s\end{aligned}

## Step 2: Calculating ratio of average lives

The ratio of the average life of a human to the average life of an extremely unstable atomic nucleus can be computed as

\begin{aligned}{}\frac{{{t_{nucleus}}}}{{{t_{human}}}} = \frac{{{{10}^{ - 22}}{\rm{ }}s}}{{2.5 \times {{10}^9}{\rm{ }}s}}\\{t_{nucleus}} = 0.4 \times {10^{ - 31}} \times {t_{human}}\end{aligned}

$${t_{nucleus}} = 4.0 \times {10^{ - 32}} \times {t_{human}}$$

Here, $$4.0$$ multiple may vary according to the average value of the human life time.

If you take$${\bf{70}} - {\bf{80}}{\rm{ }}{\bf{years}}$$, the ratio will be about $$3 \times {10^{ - 32}}$$, and if we take the human life time as $$100{\rm{ }}years$$, then the ratio will be about $$5 \times {10^{ - 32}}$$.

## Step 3: Deriving conclusions

Therefore, the life time of a human being is longer than the mean life time of an unstable nucleus by an order $${10^{31}}$$. ### Want to see more solutions like these? 