Question

1. A fossil is found to have a 14C level of 87.0% compared to living organisms. How old is the fossil?

(answer in years)

2.Iodine-131 is a radioactive isotope. After 3.00 days, 77.2% of a sample of 131I remains. What is the half-life of 131I?

(answer in days)

Answer 1

(1) All the radiactive nucleii undergo first order reactions

For a first order reaction rate constant , k = ( 2.303 /t )x log ( M_o / M)

Where

M_o = initial mass

M= Mass left after time t = 87% of M_o = 0.87 M_o

t = time = ?

k = rate constant = 0.693 / t_1/2                  Where t_1/2 = half life of C-14 = 5730 years

                         = 0.693 / 5730

                        = 1.21x10^-4 year^-1

Plug the values we get

k = ( 2.303 /t )x log ( M_o / M)

t = ( 2.303 /k)x log ( M_o / M)

t = ( 2.303 /(1.21x10^-4) )x log ( M_o / 0.87M_o)

= 1151 years

Therefore the age of the fossil is 1151 years

(2) For a first order reaction rate constant , k = ( 2.303 /t )x log ( M_o / M)

Where

M_o = initial mass

M= Mass left after time t = 77.2% of M_o = 0.772 M_o

t = time = 3.00 days

k = rate constant = ?

Plug the values we get

k = ( 2.303 /t )x log ( M_o / M)

   = (2.303/ 3) x log(M_o / 0.772M_o )

   = 0.086 day^-1

We know that half life , t_1/2 = 0.693 / k

= 0.693 / 0.086 days

= 8.03 days

Therefore the half life of I-131 is 8.03 days