Practicing Success
Practicing Success
CUET Preparation Today
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t1/2(X) = t(Y) $\Rightarrow \frac{ln2}{\lambda_X} = \frac{1}{\lambda_Y}$ $\Rightarrow \lambda_X = ln2 \lambda_Y$ $\Rightarrow \lambda_X < \lambda_Y$ Higher is the decay constant, higher is the rate of decay.
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