Q. 5.86 - An Introduction to General, Organic, and Biological Chemistry

Question

The half-life for the radioactive decay of Ce-141 is $32.5$ days. If a sample has an activity of $4.0\mu \mathrm{Ci}$ after 130 days have elapsed, what was the initial activity, in microcuries, of the sample? $(5,3,5,4)$

Step-by-Step Solution

Answer

As a result, the product is ${}_{116}{}^{296}\mathrm{Lv}$ and the complete bombardment reaction is ${}_{96}{}^{248}\mathrm{Cm}+{}_{20}{}^{48}\mathrm{Ca}\to {}_{116}{}^{296}\mathrm{Lv}$

Step 1: Introduction

Both the mass number and the atomic number must be balanced on both sides of the equation to balance the nuclear reaction. This signifies that the total mass number and atomic number of reactants and products are the same.

${}_{\%}{}^{248}\mathrm{Cm}+{}_{20}{}^{48}\mathrm{Ca}\to$

Step 2: Mass number of unknown products

The mass number of unknown products:

$248+48=A$

$A=296$

Step 3: Atomic number of unknown product

The atomic number of unknown product:

$20+96=Z$

$Z=116$

Step 4:

As a result, the product is ${}_{116}{}^{296}\mathrm{Lv}$ and the complete bombardment reaction is  ${}_{96}{}^{248}\mathrm{Cm}+{}_{20}{}^{48}\mathrm{Ca}\to {}_{116}{}^{296}\mathrm{Lv}$