Skip to content
Exercises · 13.21

Q.The half-life of 3890Sr^{90}_{38}\text{Sr} is 28 years. What is the disintegration rate of 15 mg of this isotope?

Odisha ChseTextbookSubjective· 2mImportance★★★★★est
50% · 25/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Find the number of atoms in 15 mg of Sr-90, find λ\lambda from the half-life, and multiply: A=λN≈7.88×1010A=\lambda N \approx 7.88\times10^{10} decays/s.

Step 1 — Number of atoms in 15 mg of Sr-90

N=15×10−3 g90 g/mol×NA=1.6667×10−4 mol×6.023×1023 mol−1N = \frac{15\times 10^{-3}\ \text{g}}{90\ \text{g/mol}} \times N_A = 1.6667\times 10^{-4}\ \text{mol} \times 6.023\times 10^{23}\ \text{mol}^{-1}

N=1.0038×1020 atomsN = 1.0038\times 10^{20}\ \text{atoms}

Step 2 — Decay constant

T1/2=28 yr=28×3.154×107 s=8.8312×108 sT_{1/2} = 28\ \text{yr} = 28 \times 3.154\times 10^{7}\ \text{s} = 8.8312\times 10^{8}\ \text{s}

λ=ln⁡2T1/2=0.6931478.8312×108=7.848×10−10 s−1\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693147}{8.8312\times 10^{8}} = 7.848\times 10^{-10}\ \text{s}^{-1}

Step 3 — Activity …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.