Answer: The mass of the original sample is \(12.82\times 10^{7} grams\).
Explanation:
Half-life of sample of phosphorus-32 = 14.28 days
\(\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{14.28}=0.0485 day^{-1}\)
\(N=N_o\times e^{-\lambda t}\)
\(lnN=lnN_o-\lambda t\)
N = 55 g, t= 57 days
\(\ln[55 g]=-0.0485 day^{-1}\times 57 days+ln[N_o]\)
\(\log\frac{55 g}{N_o}=2.303\times (-0.0485 day^{-1})\times 57 days\)
\(\frac{55 g}{N_o}=antilog[-6.3666]\)
\(N_o=\frac{55 g}{4.29\times 10^{-7}}=12.82\times 10^{7} grams\)
The mass of the original sample is \(12.82\times 10^{7} grams\).
