Question

a sample from a mummified bull was taken from a certain place. The sample shows that 71% of the carbon-14 still remains. how old is the sample

Answer 1

Answer:

Step-by-step explanation:

Decay of carbon - 14 is exponential in nature . It decays as follows .

$N=N_0e^{-\lambda t}$  λ is called decay constant .

λ = .693 / T where T is half life .

Half life of carbon-14 is 5700 years

λ = .693 / T

= .693 / 5700

= 12.158 x 10⁻⁵ year⁻¹

$N=N_0e^{-\lambda t}$

N = .71 N₀

$.71 N_0 =N_0e^{-\lambda t}$

$.71 =e^{-\lambda t}$

Taking ln on both sides

ln .71 = - λ t

ln .71 = - 12.158 x 10⁻⁵  t

-0.3425 =  - 12.158 x 10⁻⁵  t

t = .3425 / 12.158 x 10⁻⁵

2817  years .