Question

The half-life for the radioactive decay of calcium-47 is $4.5$ days. If a sample has an activity of $1.0\mu \mathrm{Ci}$ after 27 days, what was the initial activity, in microcuries, of the sample? $(5.3,5.4)$

Short answer

As a result, the product is${}_{112}{}^{292}\mathrm{Fl}$ The following is the whole bombardment reaction$\underset{94}{244}\mathrm{Pu}+\frac{48}{20}\mathrm{Ca}\to {}_{112}{}^{292}\mathrm{Fl}$

Step-by-step solution

Introduction

To balance a nuclear reaction, both the mass number and the atomic number must balance, which means that the total mass number and atomic number of reactants must equal the result.${}_{24}{}^{244}\mathrm{Pu}+{}_{20}{}^{+8}\mathrm{Ca}\to$

Mass number of unknown product

The mass number of unknown product:

$244+48=A$$A=292$

Unknown product's atomic number

The unknown product's atomic number is:

$20+94=Z$

$Z=114$

Bombardment reaction 

As a result, the product is ${}_{112}{}^{292}\mathrm{Fl}$The following is the whole bombardment reaction:$\underset{94}{244}\mathrm{Pu}+\frac{48}{20}\mathrm{Ca}\to {}_{112}{}^{292}\mathrm{Fl}$