Question

(a) Show that the total binding energy ${\mathit{E}}_{\mathbf{b}\mathbf{e}}$of a given nuclide is${\mathit{E}}_{\mathbf{be}}\mathbf{=}\mathit{Z}{\mathbf{∆}}_{\mathbf{H}}\mathbf{+}\mathit{N}{\mathbf{∆}}_{\mathbf{n}}\mathbf{-}\mathbf{∆}$, where, ${\mathbf{∆}}_{\mathbf{H}}$is the mass excess of ${}^{\mathbf{1}}\mathit{H}$,${\mathbf{∆}}_{\mathbf{n}}$is the mass excess of a neutron, and $\mathbf{∆}$is the mass excess of the given nuclide. (b) Using this method, calculate the binding energy per nucleon for ${}^{\mathbf{197}}\mathbf{A}\mathbf{u}$. Compare your result with the value listed in Table 42-1. The needed mass excesses, rounded to three significant figures, are ${\mathbf{∆}}_{\mathbf{H}}\mathbf{=}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{29}\mathbf{}\mathit{M}\mathit{e}\mathit{V}$, ${\mathbf{∆}}_n\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{07}\mathbf{}\mathit{M}\mathit{e}\mathit{V}$, and${\mathbf{∆}}_{197}\mathbf{=}\mathbf{+}\mathbf{31}\mathbf{.}\mathbf{2}\mathbf{}\mathit{M}\mathit{e}\mathit{V}$. Note the economy of calculation that results when mass excesses are used in place of the actual masses.

Short answer

1. The total binding energy $E_{be}$of a given nuclide is $E_{be}=Z{∆}_H+N{∆}_n-∆$.
2. The binding energy per nucleon for ${}^{197}Au$is 7.92 MeV.

Step-by-step solution

Given data

The mass excess of${}^{1}H$, ${∆}_H=+7.29MeV$

The mass excess of a neutron,${∆}_n=+8.07MeV$

The mass excess of ${}^{197}Au$,$∆=-31.2MeV$

Understanding the concept of binding energy  

The binding energy of an element is defined as the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. It can simply also be stated as the difference in mass energy between a nucleus and its nucleons. Now, by simply dividing the whole energy value by the nucleon number, we can get the required value of binding energy per nucleon.

Formulae:

The binding energy of an atom,$∆E_{be}=\left[Z{m}_H+\left(A-Z\right)m_H-M\right]c^{2}or∆m{c}^2........\left(1\right)$

Where, Zis the atomic number (number of protons), Ais the mass number (number of nucleons), $M_H$is the mass of a hydrogen atom, $M_n$is the mass of a neutron, and $M_{atom}$is the mass of an atom.

The binding energy per nucleon of an atom, $∆E_{be/nucleon}=∆E_{be}/A.......\left(2\right)$

a) Calculation of the formula of binding energy

If the masses are given in atomic mass units, then mass excesses are defined by:

$\begin{array}{l}{∆}_H=\left(m_H-1\right)c^2,\\ {∆}_n=\left(m_n-1\right)c^2,\\ ∆=(M-A)c^2,\end{array}$

This can also be written as follows:

$\begin{array}{l}{m}_H{c}^2={∆}_H-1c^2,\\ m_n{c}^2={∆}_n-1c^2,\\ m{c}^2=∆-A{c}^2,\end{array}$

Thus, substituting these equations in equation (1), we can get the binding energy of a given nuclide as follows:

$$\begin{gathered} ∆E_{be}=\left[Z\left({∆}_H+c^2\right)+(A-Z)\left({∆}_n+c^2\right)-\left(∆+c^2\right)\right] \\ =Z{∆}_H+(A-Z){∆}_n-∆ \\ =Z{∆}_H+N{∆}_n-∆,where,(A-Z)=N,numberpfneutrons........\left(3\right) \end{gathered}$$

Hence, the total binding energy $∆E_{be}$of a given nuclide is $∆E_{be}=Z{∆}_H+N{∆}_n-∆$.

b) Calculation of the binding energy per nucleon of Gold atom

Using the given data in equation (3), we can get the binding energy of the Gold atom with atomic number Z = 79 and mass number as follows:

$$\begin{gathered} ∆E_{be}=79\left(7.29MeV\right)+\left(197-79\right)\left(+8.07MeV\right)-\left(-31.2MeV\right) \\ =1560MeV \end{gathered}$$

Now, using the above value, we can get the binding energy per nucleon from equation (2) as follows:

$$\begin{gathered} ∆E_{be/nucleon}=1560MeV/197 \\ =7.92MeV \end{gathered}$$

Hence, the value of energy per nucleon is 7.92 MeV.