Question

A 0.133 pg sample of iron-59 produces 242 β-particles per second. what is the half-life of this isotope (in hours)

Answer 1

1070 hours.  
1 mole of iron-59 would mass 59 grams, so 0.133 picograms would be 0.133x10^-12 / 59 = 2.25x10^-15 moles of iron-59. Multiplying by Avogadro's number, we can determine the number of atoms of iron-59 we have, so: 2.25x10^-15 * 6.02214x10^23 = 1.35x10^9 
 Since we have 242 decays over a period of 1 second, we can divide the
number of atoms left by the original number of atoms
 (1350000000 - 242)/1350000000
 = 1349999758/1350000000
 = 0.999999820740741 
 And calculate the logarithm to base 2 of that quotient.
 ln(0.999999820740741)/ln(2)
 = -1.79259275281191x10^-7/0.693147180559945
 = -2.58616467481524x10^-7 
 The reciprocal of this number will be the half life in seconds. So
 -1/2.58616467481524x10^-7
 = -3866729.79388461 
 And dividing by 3600 (number of seconds in an hour) will give the half-life in
hours.
 -3866729.79388461 / 3600 = -1074.091609 
 So the half life in hours to 3 significant figures is 1070 hours.
 Dividing that figure by 24 gives a half life of 44.58 days which is in pretty close agreement to the official half-life of 44.495 days for iron-59.