Physics - Modern Physics Question with Solution | TestHub

No. of nuclide at time $\mathrm{t}$ is given by $$\mathrm{N}={\mathrm{N}}_{\mathrm{o}}{\mathrm{e}}^{-\lambda \mathrm{t}}$$ Where, ${\mathrm{N}}_{\mathrm{o}}=$ initial nuclide Thus this equation is equivalent to $\mathrm{y}={\mathrm{ae}}^{-\mathrm{kx}}$ So correct graph is in option (c)