What is the relationship between the decay co | vedclass

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(A) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.At half-life $t = 	au_{1/2}$,the number of nuclei remaining is $N = N_0 / 2$.Su

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$	au_{1/2} = \frac{\ln 2}{\lambda}$

Vedclass · Answer

(A) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.At half-life $t = 	au_{1/2}$,the number of nuclei remaining is $N = N_0 / 2$.Substituting these values: $\frac{N_0}{2} = N_0 e^{-\lambda 	au_{1/2}}$.$1/2 = e^{-\lambda 	au_{1/2}}$,which implies $2 = e^{\lambda 	au_{1/2}}$.Taking the natural logarithm on both sides: $\ln 2 = \lambda 	au_{1/2}$.Therefore,the relationship is $	au_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$.

What is the relationship between the decay constant $\lambda$ and the half-life $\tau_{1/2}$?

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