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Additional Exercises · 13.25

Q.A source contains two phosphorous radio nuclides 1532P^{32}_{15}\text{P} (T1/2T_{1/2} = 14.3d) and 1533P^{33}_{15}\text{P} (T1/2T_{1/2} = 25.3d). Initially, 10% of the decays come from 1533P^{33}_{15}\text{P}. How long one must wait until 90% do so?

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Write each isotope's activity as Ai(t)=Ai(0)e−λitA_i(t) = A_i(0)e^{-\lambda_i t}, use the initial 10%/90% split to fix A1(0)/A2(0)=9A_1(0)/A_2(0)=9, then solve for the time at which the ratio flips to 1/9. Result: about 208.5 days.

Let subscript 1 denote 32P^{32}\text{P} (T1/2=14.3T_{1/2}=14.3 d) and 2 denote 33P^{33}\text{P} (T1/2=25.3T_{1/2}=25.3 d).

Step 1 — Initial condition

Initially, P-33 supplies 10% of decays and P-32 supplies 90%, so:

A1(0)A2(0)=9010=9\frac{A_1(0)}{A_2(0)} = \frac{90}{10} = 9

Step 2 — Condition at the target time

We want P-33 to supply 90% of decays (P-32 the remaining 10%):

A1(t)A2(t)=1090=19\frac{A_1(t)}{A_2(t)} = \frac{10}{90} = \frac{1}{9}

Step 3 — Express the ratio's time evolution

A1(t)A2(t)=A1(0)A2(0) e−(λ1−λ2)t=9 e−(λ1−λ2)t\frac{A_1(t)}{A_2(t)} = \frac{A_1(0)}{A_2(0)}\, e^{-(\lambda_1-\lambda_2)t} = 9\, e^{-(\lambda_1-\lambda_2)t}

Setting this equal to 1/91/9:

9 e−(λ1−λ2)t=19  ⟹  e−(λ1−λ2)t=1819\, e^{-(\lambda_1-\lambda_2)t} = \frac{1}{9} \implies e^{-(\lambda_1-\lambda_2)t} = \frac{1}{81}

(λ1−λ2)t=ln⁡(81)=4ln⁡3=4.3944(\lambda_1-\lambda_2)t = \ln(81) = 4\ln 3 = 4.3944

Step 4 — Decay constants

λ1=ln⁡214.3=0.048472 day−1,λ2=ln⁡225.3=0.027393 day−1\lambda_1 = \frac{\ln 2}{14.3} = 0.048472\ \text{day}^{-1}, \qquad \lambda_2 = \frac{\ln 2}{25.3} = 0.027393\ \text{day}^{-1}

λ1−λ2=0.021079 day−1\lambda_1 - \lambda_2 = 0.021079\ \text{day}^{-1}

Step 5 — Solve for t

t=4.39440.021079=208.5 dayst = \frac{4.3944}{0.021079} = 208.5\ \text{days}

✓Final answer

t≈208.5 dayst \approx \boxed{208.5\ \text{days}}

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