Question

21.42

for the beta decay of 130TI? OBJECTIVE Determine the ages of samples of matte from radioactive decay data. 21.41 A sample of uranium ore contains 6.73 mg 238U and 3.22 ,206Pb. Assuming that all of the 206Pb arose from decay mg of the 238U and that the half-life of 23U is 4.51 × 10° years, determine the of the ore. age 21.42 A tree was cut down and used to make a statue 2300 vears ago. What fraction of the 14C that was present originally remains today? (Note: t/2 of 14C is 5730 years.) 21.43 The ratio of 87Rb to 87Sr can be used for the dating of geo- logical samples, assuming that all of the strontium came from the beta decay of rubidium (t/2 = 5.0 X 1010 years). What is the age of a rock sample if 87Sr/87RB = 0.051? 21.44 A In a living organism, the decay of 14C produces 15.3 disintegrations per minute per gram of carbon. The half- life of 14C is 5730 years. What percentage of the atoms of %3D carbon in the biosphere is 1"C? OBJECTIVE Write the nuclear equations that describe transmutations. 21.45 Complete and balance the following nuclear equations. (a) Fe +a-→ 2;p + Na +a + 5,n Tc +n (b) +in (c) 2 U+ 0 (d) Mo + (e) 250 Cf + B →+5,n 21.46 Complete and balance the following nuclear equations. 98 Department of Energy/Photo Researchers, Inc.

Answer 1

Q21.42 Time passed = 2300 years

half life of ¹⁴C = 5730 years

half lives passed = (Time passed) / (half life of ¹⁴C)

half lives passed = (2300 years) / (5730 years)

half lives passed = 0.4014

Let initial amount of ¹⁴C = 100 g

According to radioactive decay law,

final amount of ¹⁴C = (initial amount of ¹⁴C) / (2^a)

where a = half lives passed = 0.4014

Substituting the values,

final amount of ¹⁴C = (100 g) / (2^0.4014)

final amount of ¹⁴C = (100 g) / (1.32)

final amount of ¹⁴C = 75.7 g

fraction of ¹⁴C remains today = (final amount of ¹⁴C / initial amount of ¹⁴C)

fraction of ¹⁴C remains today = (75.7 g / 100 g)

fraction of ¹⁴C remains today = 0.757