# Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of 14C averaged about 0.255 Bq per gram of carbon. - Physics

Numerical

Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of 14C averaged about 0.255 Bq per gram of carbon.
(a) What fraction of carbon atoms were 14C?
(b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? The half-life of 14C is 5730 years.

#### Solution

Data: T1/2 = 5730 y

∴ lambda = 0.693/(5730 xx 3.156 xx 10^7)"s"^-1

= 3.832 × 10-12 s-1, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10-3 g,

174 decays in one hour = 174/3600 dis/s = 0.04833 dis/s in part (b) [per 500 mg]

(a) A = Nλ

∴ N = "A"/lambda = 0.255/(3.832 xx 10^-12)

= 6.654 × 1010

Number of atoms in 1 g of carbon = (6.02 xx 10^23)/12 = 5.017 xx 10^22

(5.017 xx 10^22)/(6.654 xx 10^10) = 0.7539 xx 10^12

∴ 1 14C atom per 0.7539 x 1012 atoms of carbon

∴  4 14C atoms per 3 x 1012 atoms of carbon

(b) Present activity per gram = 0.04833/(500 xx 10^-3)

= 0.09666 dis/s per gram

A0 = 0.255 dis/s per gram

Now, A(t) = "A"_0"e"^(-lambda"t")

∴ lambda"t" = 2.303 log_10 "A"_0/"A" = 2.303 log_10 (0.255/0.09666)

∴ t = (2.303 log 2.638)/(3.832 xx 10^-12) = ((2.303)(0.4213))/(3.832 xx 10^-12)

= 25.31 × 1010 s

= (25.32 xx 10^10)/(3.156 xx 10^7) = 8023 years

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 22 | Page 343