Question

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample?

Answer 1

55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g

Answer 2

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 $12.82\times 10^{7} grams$.