In the case of a radioisotope,the value of T_ | vedclass

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(B) Given that the magnitude of half-life $(T_{1/2})$ is equal to the decay constant $(\lambda)$,we have $T_{1/2} = \lambda$. We know the relationship

Vedclass · Accepted Answer

$(0.693)^{1/2}$

Vedclass · Answer

(B) Given that the magnitude of half-life $(T_{1/2})$ is equal to the decay constant $(\lambda)$,we have $T_{1/2} = \lambda$. We know the relationship between half-life and decay constant is $T_{1/2} = \frac{0.693}{\lambda}$. Substituting $T_{1/2} = \lambda$ into the equation,we get $\lambda = \frac{0.693}{\lambda}$. This implies $\lambda^2 = 0.693$. Therefore,$\lambda = \sqrt{0.693} = (0.693)^{1/2}$. Since $T_{1/2} = \lambda$,the value is $(0.693)^{1/2}$.

In the case of a radioisotope,the value of $T_{1/2}$ and $\lambda$ are identical in magnitude. The value is

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