A radioactive element has a half life of 200 days. The percentage of original activity remaining aft — Chemical Kinetics Chemistry Question
Question
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is _____. (Nearest integer) (Given: antilog 0.125 = 1.333, antilog 0.693 = 4.93)
💡 Solution & Explanation
**Step 1: Identify the radioactive decay formula** Use the equation: N(t) = N₀ × (1/2)^(t/t₁/₂) Where: - N(t) = activity remaining after time t - N₀ = original activity - t = elapsed time = 83 days - t₁/₂ = half-life = 200 days **Step 2: Calculate the exponent** t/t₁/₂ = 83/200 = 0.415 **Step 3: Simplify the decay equation** N(t)/N₀ = (1/2)^0.415 Taking logarithm: log(N/N₀) = 0.415 × log(1/2) log(N/N₀) = 0.415 × (-0.301) log(N/N₀) = -0.125 **Step 4: Convert using antilog** N/N₀ = 10^(-0.125) = 1/10^0.125 Using the given data: antilog 0.125 = 1.333 Therefore: N/N₀ = 1/1.333 = 0.75 **Step 5: Convert to percentage** Percentage remaining = 0.75 × 100 = 75% Therefore, the answer is 75.