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2. Hello
Originally Posted by jisaac
The graph of radioactive count rate N against elapsed time t is an exponential; the equation of this graph is

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?
\displaystyle \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 $\displaystyle y=\ln N_0 -\lambda t$. Acknowledged that the gradient is the coefficient which multiplies $\displaystyle t$ and that the intercept is $\displaystyle y(0)$, can you find these two constants ?

3. Originally Posted by jisaac
The graph of radioactive count rate N against elapsed time t is an exponential; the equation of this graph is

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

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

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

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

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