Using A = A0 e^{-λ t}, calculate A given A0=205862 Bq, λ=2.97×10^-6 s^-1, t=409738 s.

Using A = A0 e^{-λ t}, calculate A given A0=205862 Bq, λ=2.97×10^-6 s^-1, t=409738 s.

• A. 60,964 Bq
• B. 120,000 Bq
• C. 20,000 Bq
• D. 6,096 Bq

Answer: (A)

This question uses the radioactive decay law, where activity decreases exponentially with time as A = A0 e^{-λ t}. Plug in the given values: compute the exponent first, λ t = (2.97×10^-6 s^-1) × (409,738 s) ≈ 1.2169. The remaining fraction is e^{-1.2169} ≈ 0.296. Multiply by the initial activity: A ≈ 205,862 Bq × 0.296 ≈ 60,964 Bq. So the activity after 409,738 seconds is about 6.10×10^4 Bq.

This question uses the radioactive decay law, where activity decreases exponentially with time as A = A0 e^{-λ t}. Plug in the given values: compute the exponent first, λ t = (2.97×10^-6 s^-1) × (409,738 s) ≈ 1.2169. The remaining fraction is e^{-1.2169} ≈ 0.296. Multiply by the initial activity: A ≈ 205,862 Bq × 0.296 ≈ 60,964 Bq. So the activity after 409,738 seconds is about 6.10×10^4 Bq.