Question

The radioactive decay series of $^{226}$Ra is as follows:

The times indicated are half-lives, the units are y = years, d = days, m = minutes. The first decay, marked t above, has a value more than 300 years. A pure sample containing 152.63 mg of $^{226}$Ra (atomic mass is 226 u) was sealed and allowed to stand for 50 days. The total rate of alpha-decay from the sample was then determined by scintillation to be 30.00 GBq. What is the half-life (in years) of $^{226}$Ra? 1Bq = 1 count s$^{-1}$ and Avogadro number = 6 x $10^{23}$ In2 = 0.69, 1 year = 3,15,36,000 s, Report your answer to nearest integer

Answer 1

1187

Answer 2

We begin by noting that in the decay chain of ²²⁶Ra the following alpha‐emitters are present within 50 days:

1. ²²⁶Ra itself   (α emission)
2. ²²²Rn      (α emission; t₁/₂ = 3.825 d)
3. ²¹⁸Po      (α emission; t₁/₂ = 3.10 m)
4. ²¹⁴Po      (α emission; t₁/₂ = 19.9 m)

The later member ²¹⁰Po (α emitter) is produced via ²¹⁰Pb (t₁/₂ = 23.3 y) and so has not reached equilibrium in 50 d (its activity is only ≈0.4% of that from Ra). Thus, in a sealed sample allowed to stand 50 days, every ²²⁶Ra decay is “followed” by fast decays of its short‐lived daughters contributing 4 α decays.

Let A_Ra be the activity (in decays/s) of ²²⁶Ra. Then the measured total alpha count rate is

$$A_{\text{total}} \approx 4\,A_{\text{Ra}}.$$

We are told

$$A_{\text{total}} = 30.00~\text{GBq} = 3.0 \times 10^{10}~\text{s}^{-1}.$$

Thus,

$$A_{\text{Ra}} = \frac{3.0 \times 10^{10}}{4} = 7.5 \times 10^9~\text{s}^{-1}.$$

Next, we find the number of ²²⁶Ra atoms in the sample. Given a mass of 152.63 mg:

$$\text{Mass} = 0.15263~\text{g},\quad \text{Molar mass} = 226~\text{g/mol}.$$

Number of moles:

$$n = \frac{0.15263}{226} \approx 6.76 \times 10^{-4}~\text{mol}.$$

Using Avogadro’s number (6.0 × 10²³ mol⁻¹),

$$N = 6.76 \times 10^{-4}\times 6.0 \times 10^{23} \approx 4.06 \times 10^{20}~\text{atoms}.$$

The decay constant for Ra-226 is given by:

$$\lambda = \frac{A_{\text{Ra}}}{N} = \frac{7.5 \times 10^9}{4.06 \times 10^{20}} \approx 1.85\times10^{-11}~\text{s}^{-1}.$$

Thus, the half-life is

$$T_{1/2} = \frac{\ln2}{\lambda} = \frac{0.693}{1.85\times10^{-11}} \approx 3.74 \times 10^{10}~\text{s}.$$

Convert seconds to years (1 year = 3.1536 × 10⁷ s):

$$T_{1/2} \approx \frac{3.74 \times 10^{10}}{3.1536 \times 10^{7}} \approx 1187~\text{years}.$$

Minimal Explanation

1. Only the short‐lived daughters ²²²Rn, ²¹⁸Po, ²¹⁴Po achieve equilibrium in 50 days. Total α count = 4× activity of ²²⁶Ra.
2. ²²⁶Ra atoms calculated from 152.63 mg sample.
3. Use A = λN and T₁/₂ = 0.693/λ to get T₁/₂ ≈ 1187 years.

Final Answer

Half-life of ²²⁶Ra ≈ 1187 years.