The variation of decay rate with the number o | vedclass

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(C) According to the law of radioactive decay,the rate of decay is given by the expression:$-\frac{dN}{dt} = \lambda N$where $\lambda$ is the decay co

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$C$

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(C) According to the law of radioactive decay,the rate of decay is given by the expression:$-\frac{dN}{dt} = \lambda N$where $\lambda$ is the decay constant and $N$ is the number of active nuclei present at time $t$.This can be rewritten as:$\frac{dN}{dt} = -\lambda N$This equation represents a linear relationship between the decay rate $(R = -\frac{dN}{dt})$ and the number of active nuclei $(N)$,where the slope is negative $(-\lambda)$.Comparing this to the equation of a straight line passing through the origin,$y = mx$,where $y = \frac{dN}{dt}$ and $x = N$,the slope $m = -\lambda$.Therefore,the graph of $\frac{dN}{dt}$ versus $N$ is a straight line passing through the origin with a negative slope.Looking at the provided options,graph $A$ represents a straight line passing through the origin with a negative slope.Thus,the correct option is $A$.

The variation of decay rate with the number of active nuclei is correctly shown in which graph?

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