Answer
(a) 1 millicurie represents *3.7*10^{7}* disintegrations/sec.
Rate = $\lambda N$
$\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{2.2*10^{6}*365*24*3600} = 9.99*10^{-15} sec^{-1}$
N = number of atoms in 0.5g sample of Neptunium-237
$\frac{0.5}{237} * 6.023*10^{23} = 1.27*10^{21} atoms$
Hence, rate = $\lambda N = 9.99*10^{-15}*1.27*10^{21} = 1.27*10^{7}$ atoms/sec
$= 1.27*10^{7} disintegrations/sec$
Activity in millcurie is $1.27*10^{7}*\frac{1}{3.7*10^{7}} = 0.343$ millicurie.
(b) Decay equation is:
$Np_{93}^{237}-> \alpha_{2}^{4} + Pa_{91}^{233}$
Work Step by Step
(a) 1 millicurie represents *3.7*10^{7}* disintegrations/sec.
Rate = $\lambda N$
$\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{2.2*10^{6}*365*24*3600} = 9.99*10^{-15} sec^{-1}$
N = number of atoms in 0.5g sample of Neptunium-237
$\frac{0.5}{237} * 6.023*10^{23} = 1.27*10^{21} atoms$
Hence, rate = $\lambda N = 9.99*10^{-15}*1.27*10^{21} = 1.27*10^{7}$ atoms/sec
$= 1.27*10^{7} disintegrations/sec$
Activity in millcurie is $1.27*10^{7}*\frac{1}{3.7*10^{7}} = 0.343$ millicurie.
(b) Decay equation is:
$Np_{93}^{237}-> \alpha_{2}^{4} + Pa_{91}^{233}$
