Question 21.SE.8: How much energy is lost or gained when 1 mol of cobalt-60 un...

Analyze We are asked to calculate the energy change in a nuclear reaction.

Plan We must first calculate the mass change in the process. We are given atomic masses, but we need the masses of the nuclei in the reaction. We calculate these by taking account of the masses of the electrons that contribute to the atomic masses.

Solve

A {}_{27}^{60}Co atom has 27 electrons. The mass of an electron is 5.4858 × 10^{-4} u. (See the list of fundamental constants in the back inside cover.) We subtract the mass of the 27 electrons from the mass of the {}_{27}^{60}Co atom to find the mass of the {}_{27}^{60}Co nucleus:

\begin{aligned}& 59.933819 \,u -(27)\left(5.4858 \times 10^{-4}\,u \right) \\& \quad=59.919007\, u\, (\text{or}\,\,59.919007 \,g / mol )\end{aligned}

Likewise, for {}_{28}^{60}Ni, the mass of the nucleus is:

\begin{aligned}& 59.930788 \,u -(28)\left(5.4858 \times 10^{-4}\,u \right) \\&=59.915428 \,u (\,\text{or}\,\,59.915428 \,g / mol )\end{aligned}

The mass change in the nuclear reaction is the total mass of the products minus the mass of the reactant:

\begin{aligned}\Delta m &=\text{mass of electron}+ \text{mass}\,{}_{28}^{60}\text{Ni nucleus}- \text{mass of}\,{}_{27}^{60}\text{Co nucleus}\\&=0.00054858 \,u+59.915428 \,u -59.919007 \,u \\&=-0.003030 \,u\end{aligned}

Thus, when a mole of cobalt-60 decays,

Δm = -0.003030 g

Because the mass decreases (Δm < 0), energy is released (ΔE < 0).The quantity of energy released per mole of cobalt-60 is calculated using Equation 21.22:

E = mc²            (21.22)

\begin{aligned}\Delta E &=c^2 \Delta \,m \\&=\left(2.9979 \times 10^8 \,m / s \right)^2(-0.003030 \,g )\left(\frac{1\, kg}{1000 \,g}\right) \\&=-2.723 \times 10^{11}\frac{kg – m ^2}{s ^2}=-2.723 \times 10^{11}\,J\end{aligned}