Question

Glancing at the binding energy per nucleon plot of Figure 11.14– with its maximum around iron Z=26––it might naively be argued that fission would be favorable for any nucleus of about Z = 52 or higher. Show that the Tellurium – 130 broke into two iron- 56 nuclei, freeing the excess neutrons, the total binding energy would actually decreases.Explain what is wrong with the naïve argument.

Short answer

The binding energy decreases.

The naïve argument assumed that all the nucleons would be bound in the more stable partial with higher binding energy per nucleon but ignored the free neutrons.

Step-by-step solution

Given data

The element is Tellurium-$130$ .

Concept of Energy released during Electron Capture 

During electron capture, the amount of energy released depends on the mass difference between the parent and the daughter nucleus.

Which is:

$\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{parent}}}\mathbf{-}{\mathbf{m}}_{\mathbf{\text{daughter}}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$.

Where${\mathbf{m}}_{\mathbf{\text{parent}}}$is the atomic mass of the parent nucleus, and${\mathbf{m}}_{\mathbf{\text{daughter}}}$is the atomic mass daughter nucleus.

Determine the Kinetic Energy

The fission of Tellurium-$130$into two iron - $56$ would be:

${}_{52}^{130}\text{Te}{\to }_{26}^{56}\text{Fe}{+}_{26}^{56}\text{Fe}+18\mathrm{n}$.

The $Q$ factor is calculated as:

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{parent}}-{\mathrm{m}}_{\text{daughter}}){\mathrm{c}}^2\\ =(129.906229\mathrm{u}-2(55.934939\mathrm{u})-18(1.008665\mathrm{u}\left)\right)(931.5\text{\hspace{0.17em}MeV/}\mathrm{u})\\ =-111.43\text{\hspace{0.17em}MeV}\end{array}$

Thus, The binding energy decreases.

The naïve argument assumed that all the nucleons would be bound in the more stable partial with higher binding energy per nucleon but ignored the $18$free neutrons.