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The isotope TI-208 undergoes beta decay with a half-life of 3.1 min. rnwhat is the decay constant for this process?rn4.47 min-¹rn0.031 min-¹rn0.224 min-¹rn2.15 min-¹
Accepted Answer
The decay constant (λ) is related to the half-life (t½) by the equation: λ = ln(2) / t½.
In this case, t½ = 3.1 min. Therefore, λ = ln(2) / 3.1 min ≈ 0.224 min⁻¹.
So the correct answer is 0.224 min⁻¹.
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