44. (a) The rate at which Radium-226 is decaying is

ln 2 M (ln 2)(1.00 mg)(6.02 × 1023/mol)
R = N = = =3.66 × 107 s-1 .
T1/2 m (1600 y)(3.15 × 107 s/y)(226 g/mol)
(b) Since 1600 y 3.82 d the time required is t 3.82 d.
(c) It is decaying at the same rate as it is produced, or R =3.66 × 107 s-1.
(d) From RRa = RRn and R = N =(ln 2/T1/2)(M/m), we get


T1/2Rn mRn
MRn = MRa
T1/2Ra mRa
(3.82 d)(1.00 × 10-3 g)(222 u)
=
(1600 y)(365 d/y)(226 u)
= 6.42 × 10-9 g .