16: 130 I decays by emission of beta particles to form stable 130 Xe. A 3.00 g i

Question

16: 130 I decays by emission of beta particles to form stable 130 Xe. A 3.00 g iodine sample containing some I-130 was recorded as having 2835 disintegrations per min. k = 0.00094 min -1. How many radioactive I-130 atoms are present in the sample?

17: The radioactive decay of a sample containing an unknown radioactive isotope produced 8633 disintegrations per minute. 3.74 days later, the rate of decay was found to be 2602 disintegrations per minute. Calculate the half-life in days for the unknown radioactive isotope.

18: A certain radioactive element has a half life of 7641 years. How much of a 2.62 g sample is left after 12706 years?

19: The half-life of carbon-14 is 5,730 years. An artifact produces 5.2 disintegrations of 14C per minute per gram of carbon in the sample. Estimate the age of this sample assuming that its original radioactivity was 15.3 disintegrations per minute per gram of carbon.

20: What is the energy released when 1.413 g are converted to energy? Express your answer in joules. For self-consistent units you must use kg because 1 J = 1 kg m2/s2.

21: What is the mass defect of 27Al in u? The mass of 1 atom of Al-27 is 26.9815 u.

22: What is the nuclear binding energy per mole if the mass defect of a certain isotope is 0.2899 u per atom or 0.2899 g per mol? Express your answer in joules. Remember to use self-consistent units.

Explanation / Answer

16.

   Radio active diintrigration is define is propotional to number atom present at that time. If atom present at that time is N and disintrrigation constant is K

Disntrigation per min = K * N

N = (2835 / .00094 ) = 3.015 *106 atom present .

17. Let at intial number of atom is N1 when sample disintrigate 8633 per min now ater 3.74 (t) day atom number became N2 at that time disintrigation 2602.

since disintrigation is proposional to number of atom present so

N1 / N2 = (8633 / 2602 ) =3.317

now Disintrigation constant K = ln (N1/ N2) / t

                                           K = 0.3206 Day-1

Half life t1/2 = ln2 / k = 2.16 day

18. Number of half life is over = 12706 / 7641 = 1.66

so, fraction of sample remaining = (0.5)1.66= 0.315

amount of sample remaining = (2.62 * .315) = 0.827 gm

19. Disintrigation constant K = ln2 / half life time = 1.23 * 10-04 year -1

now ration of atom frominitial and now (N1/ N2) = 15.3 / 5.3 = 2.886

now K = ln (N1/ N2) / t

t = 51211.2 years is the age of sample.

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