Quantitative analysis of decay (HL)

Answer

$N = N_0 e^{-\lambda t}$ — exponential decay of the un-decayed nuclei.

Answer

$A = A_0 e^{-\lambda t}$, and $A = \lambda N$. Activity follows the same curve as N.

Answer

The **probability per unit time** that a given nucleus decays. Unit: **s⁻¹** (or day⁻¹, hour⁻¹).

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$\lambda = \dfrac{\ln 2}{t_{1/2}}$ (ln 2 ≈ 0.693).

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Because $A = \lambda N$ — activity is just λ times N, so both decay by the same factor $e^{-\lambda t}$.

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$\left(\tfrac{1}{2}\right)^{n}$ — e.g. after 3 half-lives, ⅛.

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$\lambda = \ln 2 / 8.0 = 0.087$ day⁻¹.

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24 days = 3 half-lives, so $(½)^3 = ⅛ = 0.125\,N_0$.

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Each nucleus is **very likely** to decay each second, so the **half-life is short**.

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The **activity A** (decays per second, in **Bq**), not N directly.

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t must use the **same time unit** as λ (λ in day⁻¹ ⇒ t in days).

Answer

Whole n half-lives ⇒ use $(½)^n$; any other time ⇒ use $e^{-\lambda t}$.