Page:Calculus Made Easy.pdf/182

It is found by putting this equation in its logarithmic form, namely,

$\log 0.5=-0.02\times l\times \log \epsilon$,

which gives

$l={\frac {-0.3010}{-0.02\times 0.4343}}=34.7$ centimetres nearly.

(3) The quantity $Q$ of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity $Q_{0}$ of the substance by the relation $Q=Q_{0}\epsilon ^{-\lambda t}$, where $\lambda$ is a constant and $t$ the time in seconds elapsed since the transformation began.

For “Radium $A$,” if time is expressed in seconds, experiment shows that $\lambda =3.85\times 10^{-3}$. Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

We have $0.5=\epsilon ^{-0.00385t}$.

$\log 0.5=-0.00385t\times \log \epsilon ;$
and $\quad \quad \quad t=3$ minutes very nearly.

 

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Exercises XIII. (See page 260 for Answers.)

(1) Draw the curve $y=b\epsilon ^{-{\frac {t}{T}}}$; where $b=12$, $T=8$, and $t$ is given various values from $0$ to $20$.

(2) If a hot body cools so that in $24$ minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to $1$ per cent. of the original excess.