Question

At any one time, the dominant naturally occurring radioisotopes in the human body are 14.0 mg of Potassium-40 (half-life 1.25 X 109 years) and 16.5 ng of Carbon-14 (half-life   5730 years). Calculate the body’s natural radioactivity in becquerels (Bq).

Answer 1

rogn of Que numberno. of atoms present in potassium = 6.022 x 100 igm of potassium = 61022x10²3 atoms in 40 no. of 40 noi of atoms in 14 mg (14x10 3 gm) of potassium N = 61022x10 23 x 14x103 N_20141020 half life of potassium - 40 = 1.25x 109 years Typ = 136125 years Activity of 14mg of potassium in human body R,= 0.693 x N T112 R = 0.693 x 211x1020 136.25 R = 0.01068x1020 Ri= 106810 181014 decay, second = 10681. 1x 104 Becquerel for carbon-ly nor of atoms present in 14 gm of Carbon = 6.622x102 ho. of atomo present in Igm of Carbon- 6.022x10 no. of atome present in 16.5ng ( 16.5 x 10 9 g) of carbon = 6:02241023 x 1615x10g N = 7:09x104

half life of Carbon- 14 = 5730 years Til = 5730 years wity of 1615ng of carbon in human body R = 0.693 xN' AC Tiz 01693 x 7109x109 5730 = 0.0008 58x169 disintegrations) - 8.58x10' Becquerel (BG a second Body has natural radioactivity of for potassium - Too 1068111x10 Bq for carbon = 0.00086 x 10 Bq or 8.58x10° BE [ Total radioactivity - 1068101 + 0.000867x 1014 Bq 10681.10086x 1014 Bq ł 1068111 X 10" Bq 1