A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C14 is 5730 years, then age of the wooden piece placed in the museum is approximately :

Question

A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C14 is 5730 years, then age of the wooden piece placed in the museum is approximately : (A) 10439 years (B) 13094 years (C) 19039 years (D) 39049 years

Tardigrade Admin · Accepted Answer

use, A =( A 0/2 n ), n =( t / T )=( t /5730) 2=(20/2 n ), ( t / T ) log 2= log 10 or t =(5730/0.3010)=19039 year

Q. A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C¹⁴ is 5730 years, then age of the wooden piece placed in the museum is approximately :

10439 years

13094 years

19039 years

39049 years

use, $A = \frac{A _{0}}{2 ^{n}}, n = \frac{t}{T} = \frac{t}{5730}$
$2 = \frac{20}{2 ^{n}}, \frac{t}{T} lo g 2 = lo g 10$
or $t = \frac{5730}{0.3010} = 19039$ year

Q. A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C14 is 5730 years, then age of the wooden piece placed in the museum is approximately :