Some radionuclides decay by capturing one of their own atomic electrons, a K-shell electron, say. An example is

${}^{\mathbf{49}}\mathbf{V}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}{\mathbf{\to }}^{\mathbf{49}}\mathbf{Ti}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{331}\mathbf{d}$

Show that the disintegration energy Qfor this process is given by

$\mathbf{Q}\mathbf{=}\left(m_v-m_{\mathrm{Ti}}\right){\mathbf{c}}^{\mathbf{2}}\mathbf{-}{\mathbf{E}}_{\mathbf{K}}$

where,${\mathbf{m}}_{\mathbf{v}}$and${\mathbf{m}}_{\mathbf{Ti}}$are the atomic masses of${}^{\mathbf{49}}\mathbf{V}$and${}^{\mathbf{49}}\mathbf{Ti}$, respectively, and${\mathbf{E}}_{\mathbf{K}}$is the binding energy of the vanadium K-shell electron. (Hint:Putandas the corresponding nuclear masses and then add in enough electrons to use the atomic masses.)

Half-life of the decay process${\mathrm{T}}_{1/2}=331\mathrm{d}$ ,

For nuclear fission of radioactive nuclide, the radionuclide needs to disintegrate by releasing some energy change. Again, nuclides with nuclear radii have nuclear mass acting into the disintegrating energy. But, we know that the sum of nuclear mass and the electronic mass which is the sum of total electron mass gives us the atomic mass. Thus, considering this concept, we can get the disintegration energy that is in the form of atomic masses of the nuclides.

Assuming the neutrino has negligible mass, we can get the energy of the decay process (using the predicted mass) as follows:

$∆{\mathrm{mc}}^2=\left({\mathbf{m}}_{\mathrm{Ti}}-{\mathbf{m}}_{\mathrm{v}}-{\mathrm{m}}_{\mathrm{e}}\right){\mathrm{c}}^2$

where, ${\mathbf{m}}_{\mathrm{Ti}}$is the nuclear mass of titanium and ${\mathbf{m}}_v$is the nuclear mass of vanadium.

Now, since vanadium has 23 electrons (see Appendix F and/or G) and titanium has 22 electrons, we can add and subtractto the above expression and obtain the following equation of the energy as:

$$\begin{gathered} ∆{\mathrm{mc}}^2=\left({\mathbf{m}}_{\mathbf{Ti}}+22{\mathrm{m}}_{\mathrm{e}}-{\mathbf{m}}_{\mathrm{v}}-23{\mathrm{m}}_{\mathrm{e}}\right){\mathrm{c}}^2 \\ =\left({\mathrm{m}}_{\mathrm{Ti}}-{\mathrm{m}}_{\mathrm{v}}\right){\mathrm{c}}^2 \end{gathered}$$

The above final expression involves the atomic masses of the nuclidesand that this assumes (due to the way they are usually tabulated) the atoms are in the ground states (which is certainly not the case here, as we discuss below).

Now, it can be seen that the atom is left in an excited (high energy) state due to the fact that an electron was captured from the lowest shell (where the absolute value of the energy$E_k$, is quite large for large Z ). To a very good approximation, the energy of theK-shell electron in Vanadium is equal to that in Titanium (where there is now a “vacancy” that must be filled by a readjustment of the whole electron cloud), and then the disintegration energy can be given as follows:

$$\begin{gathered} Q=-\left(∆m{c}^2\right)-E_K \\ =\left(m_v-m_{Ti}\right)c^2-E_K \end{gathered}$$

Hence, the disintegration energy Q for the decay process of vanadium-59 to titanium-49 is$Q=\left(m_v-m_{Ti}\right)c^2-E_K$ where,$m_v$and$m_{Ti}$are the atomic masses of vanadium and titanium and$E_K$is the binding energy of the vanadium K-shell electron.