StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A radioactive sample has a half-life of 40 seconds. Its activity 80 seconds after the start is measured to be 6.932 * 10¹⁸ dps. The total energy released during this 80-second interval is 6 * 10⁸ J (given ln2 = 0.6932). Which of the following are correct?

  1. The initial number of atoms in the sample is 1.6 * 10²⁰
  2. The initial number of atoms in the sample is 1.6 * 10²¹
  3. Energy released per disintegration is 5 * 10⁻¹³ J
  4. Energy released per disintegration is 5/3 * 10⁻¹³ J

Correct answer: The initial number of atoms in the sample is 1.6 * 10²¹

Solution

lambda = 0.6932/40 = 0.01733 s⁻¹. Activity after 2 half-lives: A(80) = lambda*N0/4. So N0 = 4*A(80)/lambda = 4 * 6.932e18 / 0.01733 = 1.6e21. Total decays = 3N0/4 = 1.2e21. Energy per decay = 6e8/1.2e21 = 5e-13 J. Options B and C are correct.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →