Question

Using the data in Table 19.1 to predict which is more stable: four protons, four neutrons and four electrons organized as two ${}_2{}^4\mathbf{He}$ or as one ${}^{\mathbf{8}}\mathbf{Be}$. What is the mass difference?

Short answer

Since the mass of two${}_{\text{2}}{}^{\text{4}}\text{He}$ atoms is less than the mass of one${}_4{}^8\text{Be}$ atom,${}_{\text{2}}{}^{\text{4}}\text{He}$ are more stable.

The mass difference is $9.9\times {10}^{-5}\text{u}$.

Step-by-step solution

Determining the stability

We will take the help of Table 19.1 to predict the total mass of two${}_{\text{2}}{}^{\text{4}}\text{He}$ atoms.

Mass of 1${}_{\text{2}}{}^{\text{4}}\text{He}$ atom =4.002603 units

Mass of two${}_{\text{2}}{}^{\text{4}}\text{He}$ atoms = $2\times 4.002603=8.005206\text{u}$

Mass of 1${}_4{}^8\text{Be}$ atom = 8.005305 u

Since, the total mass of two${}_{\text{2}}{}^{\text{4}}\text{He}$ is less than one${}_4{}^8\text{Be}$ atom,${}_{\text{2}}{}^{\text{4}}\text{He}$ are more stable.

Mass difference

The mass difference will be calculated by subtracting the mass of two${}_{\text{2}}{}^{\text{4}}\text{He}$ atoms from the mass of one${}_4{}^8\text{Be}$atom.

$$\begin{gathered} \Delta m=8.0053\text{05u - 8.005206} \text{u} \\ =9.9\times {10}^{-5}\text{u} \end{gathered}$$