At some instant, the number of radioactive atoms in a sample is $N_0$ and after time $t$, the number decreases

Question

At some instant, the number of radioactive atoms in a sample is $N_0$ and after time $t$, the number decreases to $N$. It is found that the graphical representation $\ln N$ versus $t$ along the $y$ and $x$ axis respectively is a straight line. Then the slope of this line is:

Accepted Answer

1. **Identify the Decay Law:** Radioactive decay follows first-order kinetics. The number of undecayed nuclei $N$ at time $t$ is given by the equation $N = N_0 e^{-\lambda t}$, where $\lambda$ is the decay constant .
2. **Linearize the Equation:** To interpret the graph, take the natural logarithm ($\ln$) of both sides:
   $$ \ln N = \ln(N_0 e^{-\lambda t}) $$
   $$ \ln N = \ln N_0 + \ln(e^{-\lambda t}) $$
   $$ \ln N = -\lambda t + \ln N_0 $$
3. **Compare with Straight Line Equation:** This equation is in the form $y = mx + c$, which represents a straight line. Matching the terms:
   - $y = \ln N$
   - $x = t$
   - $m$ (slope) = $-\lambda$
   - $c$ (intercept) = $\ln N_0$
4. **Conclusion:** The slope of the graph of $\ln N$ versus $t$ is $-\lambda$ .

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