Q. In a radioactive material the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is λ, then :
A) R₁ = R₂ e^(−λ(t₁ − t₂))
B) R₁ = R₂ e^(λ(t₁ − t₂))
C) R₁ = R₂ e^(t₂/t₁)
D) R₁ = R₂
Explanation
[Solution from source] Activity decays as R = R₀e^(−λt), so R₁ = R₀e^(−λt₁) and R₂ = R₀e^(−λt₂). Dividing, R₁ = R₂e^(−λ(t₁ − t₂)); since t₁ < t₂ the exponent is positive and R₁ > R₂, as expected for the earlier time. The form with e^(λ(t₁ − t₂)) would make the earlier activity smaller. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.