NTA Abhyas JEE Main2020PhysicsNuclear PhysicsPractice
The β - activity of a sample of C O 2 prepared from a contemporary wood gave a count rate of 25.5 counts per minute ( cpm ) . The same mass of C O 2 from an ancient wooden statue gave a count rate of 20.5 cpm under the same conditions. If the half life of 14 C is 5770 years , then the age of the statue is close to [Take log 10 255 205 ≈ 0 . 095 ]
Options
- A1822 years
- B182   years
- C822   years
- D18220   years
Correct answer
A. 1822 years
Step-by-step solution
r = 20.5 c p m , r 0 = 25.5 c p m ∵ r 0 ∝ N 0 a n d r ∝ N ∴ r 0 r = N 0 N Also, t = 2 .303 λ log N 0 N = 2 .303 λ log r 0 r t = 2 .303 × 5770 0 .693 log 25 .5 20 .5 = 1822 years