# Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 - Physics

Numerical

Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10−12per second.

#### Solution

Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,

("A"("t"))/"A"_0 = 12.3/15.3, λ = 3.839 x 10-12 per second

A(t) = A0e-λt  ∴ "e"^(lambda"t") = "A"_0/"A"

∴ lambda"t" = log_"e"("A"_0/"A")

∴ t = 2.303/lambda log_10 ("A"_0/"A")

= 2.303/(3.839 xx 10^-12)log_10(15.3/12.3)

= (2.303 xx 10^12)/3.839(log 15.3 - log 12.3)

= (2.303 xx 10^12)/3.839 (1.1847 - 1.0899)

= ((2.303)(0.0948))/3.839 xx 10^12s

= 5.687 x 1010 s

= (5.687 xx 10^10 "s")/(3.156 xx 10^7 "s per year")

= 1802 years

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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 16 | Page 343