BITSAT2019PhysicsNuclear PhysicsActual
A radioactive sample of half life 23 . 1 days is disintegrating continuously. The percentage decay of the sample between the 15 th and 16 th day will be [Take, e 0 . 03 = 1 . 03 ]
Options
- A5   %
- B1   %
- C2 . 9   %
- D3 . 5   %
Correct answer
C. 2 . 9   %
Step-by-step solution
Given, half life T 1 / 2 = 23 . 1 days If N 0 be the initial amount, then active nuclei at t = 15 days N 1 = N 0 e - 15 λ       . . . i Active nuclei at t = 16 days N 2 = N 0 e - 16 λ       . . . ii Disintegration constant, λ = 0 . 693 T 1 / 2 = 0 . 693 23 . 1 = 0 . 03 ∴   % decay in 15 th to 16 th days = N 1 - N 2 N 1 × 100 = N 0 e - 15 λ - N 0 e - 16 λ N 0 e - 15 λ × 100 = 1 - e - λ × 100 = 1 - e - 0 . 03 × 100 = 1 - 1