An$\mathit{\alpha }$particle (${}^{\mathbf{4}}\mathit{H}\mathit{e}$nucleus) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an$\mathit{\alpha }$particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and neutron mass.

$$\begin{gathered} {}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00260}\mathit{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{2}}\mathit{H}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{01410}\mathit{u} \\ {}^{\mathbf{3}}\mathit{H}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{01605}\mathit{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{1}}\mathit{H}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00783}\mathit{u} \\ \mathit{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00867}\mathit{u} \end{gathered}$$

• The atomic mass of helium,$m_{He}=4.00260u$
• The atomic mass of tritium,$m_t=3.01605u$
• The atomic mass of deuteron,$m_d=2.01410u$
• The atomic mass of the hydrogen,$m_H=1.00883u$
• The atomic mass unit of neutron, $m_n=1.00867u$

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. Now, using this energy value and dividing it by the number of nucleons of the isotope, we can get the binding energy per nucleon of the given nuclide. For each case, the removal of an atom refers to the production of the binding energy of the atom, thus using the electron conservation law, you can write the respective equation for each given situation. Now, using the given data and equating them as per the conservation equation, we can get our required energy value.

Formulae:

The binding energy of an atom,

$∆E_{be}=\left[Z{m}_H+\left(A-Z\right)m_H-M_{atom}\right]c^{2}or∆m{c}^2$ ….. (i)

Where, Z is the atomic number (number of protons), A is 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,

$∆E_{be/nucleon}=∆E_{be}/A$ ….. (ii)

The energy of an atom,

$E=m{c}^2$ ….. (iii)

Here, m is the mass and c is the speed of light.

The first step is to add energy to produce${}^4\mathrm{He}\to \rho {+}^3\mathrm{H}$, which — to make the electrons “balance” — may be rewritten as${}^4\mathrm{He}{\to }^1\mathrm{H}{+}^3\mathrm{H}$.

Thus energy needed to remove a proton can be given using the equations (i) and (iii) and the above energy equation as follows:

$\begin{array}{l}∆E_1=(m_t+m_H-m_{He})c^2\\ =(3.01605u+1.00783u-4.00260u)\left(931.5MeV/u\right)\\ =19.8MeV\end{array}$

Hence, the value of the energy is 19.8 MeV.

The second step is to add energy to produce${}^4\mathrm{He}\to n{+}^3\mathrm{H}$. The energy needed to remove a neutron can be given using the equations (i) and (iii) and the above energy equation as follows:

role="math" localid="1661921745721" $\begin{array}{l}∆E_2=(m_d+m_n-m_t)c^2\\ =(2.01410u+1.00867u-3.01605u)\left(931.5MeV/u\right)\\ =6.26MeV\end{array}$

Hence, the value of the energy is 6.26 MeV .

The third step:${}^{2}H\to p+n$, which — to make the electrons “balance” — may be rewritten as${}^{2}H\to {}^{1}H+n$. The energy required for separating a proton and a neutron from tritium atom can be given using equations (i) and (iii) as follows:

$\begin{array}{l}∆E_3=(m_H+m_n-m_d)c^2\\ =(1.00783u+1.00867u-2.01410u)\left(931.5MeV/u\right)\\ =2.23MeV\end{array}$

Hence, the value of the energy is 2.23 MeV.

The total binding energy from the energy values of part (a), (b) and (c) calculations can be written as follows:

$\begin{array}{l}∆E_{be}=∆E_1+∆E_2+∆E_3\\ =19.8MeV+6.26MeV+2.23MeV\\ =28.3MeV\end{array}$

Hence, the value of the energy is 28.3 MeV.

The binding energy per nucleon of the helium atom with mass number (or nucleons) A = 4 can be calculated using the given data in equation (ii) as follows:

$$\begin{gathered} ∆E_{be/nucleon}=28.3MeV/4 \\ =7.07MeV \end{gathered}$$

Hence, the value of the energy per nucleon is 7.07 MeV.

No, the energy answers of the parts (a), (b) and (c) do match with the energy values.