• Jun 23rd 2008, 12:08 PM
jisaac
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• Jun 23rd 2008, 12:39 PM
flyingsquirrel
Hello
Quote:

Originally Posted by jisaac
The graph of radioactive count rate N against elapsed time t is an exponential; the equation of this graph is
http://i276.photobucket.com/albums/k...saac1234/7.png
where N0 and λ are positive constants.

A graph of ln N against t will be a straight line of gradient (something) and intercept of (something) on the vertical axis.

what is the gradient and intercept?

\begin{aligned}
\ln N&=\ln\left(N_0 \mathrm{e}^{-\lambda t}\right)\\
&=\ln N_0 + \ln \left( \mathrm{e}^{-\lambda t}\right)\\
&=\ln N_0-\lambda t
\end{aligned}

so the equation of the straight line is $y=\ln N_0 -\lambda t$. Acknowledged that the gradient is the coefficient which multiplies $t$ and that the intercept is $y(0)$, can you find these two constants ?
• Jun 23rd 2008, 12:40 PM
Jhevon
Quote:

Originally Posted by jisaac
The graph of radioactive count rate N against elapsed time t is an exponential; the equation of this graph is
http://i276.photobucket.com/albums/k...saac1234/7.png
where N0 and λ are positive constants.

A graph of ln N against t will be a straight line of gradient (something) and intercept of (something) on the vertical axis.

what is the gradient and intercept?

thanks!

$N = N_0 e^{- \lambda t}$

$\Rightarrow \ln N = \ln (N_0 e^{- \lambda t})$

$\Rightarrow \ln N = \ln N_0 + \ln (e^{- \lambda t})$

$\Rightarrow \ln N = - \lambda t + \ln N_0$

now remember the slope-intercept form of the equation of a line. i leave the rest to you