Question

A. An unknown radioactive substance has a half-life of 3.20hours . If 10.6g of the substance is currently present, what mass A0 was present 8.00 hours ago?

B. Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 45.0{\rm \\%} of an Am-241 sample to decay?

C. A fossil was analyzed and determined to have a carbon-14 level that is 20{\rm \\%} that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

Answer 1

A) t_1/2 = 3.20hours

     A= 10.6g

    A0 = ?

    t= 8 hours

t= n t_1/2

8 = n x 3.2

n= 2.5

A= A0/2ⁿ

10.6 = A0/2^2.5

A₀ = 59.96g   (approximately 60g)

B) t_1/2 = 432 years

      for radio active decay disntigration constant = 0.693/ t_1/2

                                                                               = 0.693/432

                                                                              = 1.604 x 10^-3 year^-1

now   t= (2.303/) log (A0/A)

            = 2.303 x log (100/45) / 1.604 x 10^-3

             = 497.9 year

C)

t_1/2 = 5730 years

      for radio active decay disntigration constant = 0.693/ t_1/2

                                                                               = 0.693/5730

                                                                              = 1.2 x 10^-4 year^-1

t= (2.303/) log (100/20)

            = 2.303 x log (100/20) / 1.2 x 10^-4

             = 13,414 years

fossil age =13,414 years