91Ó°ÊÓ

1. First case,${}^{238}∪$nucleus emits an alpha particle.
2. Second case,${}^{238}∪$ nucleus a sequence of neutron, proton, neutron and proton.
3. Atomic mass of the particles is given.

The uranium nucleus undergoes fission through alpha, beta, or gamma decay to turn into a more stable form from the present nucleus. Considering the conservation of the electronic charge and mass of the particles, we can write an equation for each condition in the given problem. Now, using the binding energy condition, we can get the energy released in each case using the given atomic masses.

Formula:

The binding energy of an atom,$\mathbf{∆}{\mathit{E}}_{\mathbf{b}\mathbf{e}}\mathbf{=}\mathbf{∆}\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}{\mathit{c}}^{\mathbf{2}}\mathbf{=}\mathbf{931}\mathbf{.}\mathbf{5}\mathbf{}\mathit{M}\mathit{e}\mathit{V}$ (i)

Where,$\mathbf{∆}\mathit{m}$is the mass difference between the nuclei

(a)

The nuclear reaction for the condition of emitting an alpha particle is given by an alpha decay of the uranium nucleus${}^{238}∪$ as:

${}^{238}U→{}^{234}Th+{}^{4}He$

Thus, the energy released in this condition by the fission reaction can be given suing the given data in equation (i) as follows:

$$\begin{gathered} ∆\mathrm{E}=\left({\mathrm{m}}_{\mathrm{U}}-{\mathrm{m}}_{\mathrm{m}}-{\mathrm{m}}_{\mathrm{He}}\right){\mathrm{c}}^2 \\ =\left(238.05079\mathrm{u}-234.04363\mathrm{u}-4.00260\mathrm{u}\right)\left(931.5\mathrm{MeV}/\mathrm{u}\right) \\ =4.25\mathrm{MeV} \end{gathered}$$

Hence, the released energy is 4.25 MeV.

(b)

As per the problem, the reaction series is given as follows:

$$\begin{gathered} {}^{238}\mathrm{U}→{}^{237}\mathrm{U}+\mathrm{n} \\ {}^{237}\mathrm{U}→{}^{236}\mathrm{Pa}+\mathrm{p} \\ {}^{236}\mathrm{Pa}→{}^{235}\mathrm{Pa}+\mathrm{n} \\ {}^{235}\mathrm{Pa}→{}^{234}\mathrm{Th}+\mathrm{p} \end{gathered}$$

The net energy released in this reaction chain series can be given using equation (i) as follows:

$$\begin{gathered} ∆E=\left(m_{{238}_U}-m_{{237}_U}-m_n\right)c^2+\left(m_{{237}_{Pa}}-m_{{236}_{Pa}}-m_p\right)c^2+\left(m_{{236}_{Pa}}-m_{{235}_{Pa}}-m_n\right)c^2+\left(m_{{235}_U}-m_{{234}_{Th}}-m_p\right)c^2 \\ =\left(m_{{238}_U}-2m_n-2m_p-m_{{234}_{Th}}\right)c^2 \\ =\left(238.05079u-2\left(1.00867u\right)-2\left(1.00783u\right)-234.04643\right)\left(931.5MeV/u\right) \\ =-24.1MeV \end{gathered}$$

Hence, the value of the energy released is 24.1 MeV.

(c)

The binding energy of the alpha particle (having 2 protons and 2 neutrons) can be given as the mass excess due to the protons and neutrons compared to that of the mass of the helium or the alpha particle. Thus, it can be written using equation (i) as:

$∆E_{He}=\left|\left(2m_n+2m_p-m_{He}\right)c^2\right|$

This leads us to conclude that the binding energy of the$\mathrm{Î\pm }$ particle is the calculated difference of the numbers of the emitted particles of the parts (a) and (b).

(d)

Thus, the binding energy of theparticle is given using equation (i) and the given data as follows:

$$\begin{gathered} ∆E_{\mathrm{Î\pm }}=\left|\left(2m_n+2m_p-m_{He}\right)c^2\right| \\ =\left|\left(2\left(1.00866u\right)+2\left(1.00783u\right)-(4.00260u)\right)(931.5MeV/u)\right| \\ =\left|-24.1MeV-4.25MeV\right| \\ =28.3MeV \end{gathered}$$

Hence, the binding energy of the alpha particle is 28.3 MeV.