In an exponential decay, if the natural logarithm of the ratio of intact nuclei $(n)$ at time $t$ to the intact nuclei at time $t = 0$ $(n_{0})$ is plotted against time, what does the slope of the graph corresponds to:

- λ

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- β

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_{γ}

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all of the above

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Solution

The correct option is C $- λ$
Decay of nuclei follows first-order kinetics.
$n (t) = n_{0} e^{- λ t}$
Taking natural log on both sides
$ln n = ln n_{0} - λ t$
$\Rightarrow ln n - ln n_{0} = - λ t$
$ln (\frac{n}{n_{0}}) = - λ t$
Comparing this equation with $y = m x$
$∴$ Slope of the required graph is $- λ$

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In an exponential decay, if the natural logarithm of the ratio of intact nuclei (n) at time t to the intact nuclei at time t=0 (n0) is plotted against time, what does the slope of the graph corresponds to:

The correct option is C −λDecay of nuclei follows first-order kinetics.n(t)=n0e−λtTaking natural log on both sideslnn=lnn0−λt⇒lnn−lnn0=−λtln(nn0)=−λtComparing this equation with y=mx∴Slope of the required graph is −λ

In an exponential decay, if the natural logarithm of the ratio of intact nuclei $(n)$ at time $t$ to the intact nuclei at time $t = 0$ $(n_{0})$ is plotted against time, what does the slope of the graph corresponds to: