The radionuclide ${}^{\mathbf{11}}\mathbf{C}$decays according to

${}^{\mathbf{11}}\mathbf{C}\mathbf{\to }{}^{\mathbf{11}}\mathbf{B}\mathbf{+}{\mathbf{e}}^{\mathbf{+}}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{3}$

The maximum energy of the emitted positrons is 0.960 MeV. (a) Show that the disintegration energy Qfor this process is given by

role="math" localid="1661759171201" $\mathit{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{C}}\mathbf{-}{\mathbf{m}}_{\mathbf{B}}\mathbf{-}\mathbf{2}{\mathbf{m}}_{\mathbf{e}}\mathbf{)}{\mathit{c}}^{\mathbf{2}}$

Where${\mathit{m}}_{\mathbf{C}}$and${\mathit{m}}_{\mathbf{B}}$are the atomic masses of${}^{\mathbf{11}}\mathit{C}$and${}^{\mathbf{11}}\mathit{B}$, respectively, and${\mathit{m}}_{\mathbf{e}}$is the mass of a positron. (b) Given the mass values${\mathit{m}}_{\mathbf{C}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{011}\mathbf{}\mathbf{}\mathbf{434}\mathbf{}\mathbf{}\mathbf{u}$,${\mathit{m}}_{\mathbf{B}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{009}\mathbf{}\mathbf{}\mathbf{305}\mathbf{}\mathbf{}\mathbf{u}$and${\mathit{m}}_{\mathbf{e}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{000}\mathbf{}\mathbf{}\mathbf{548}\mathbf{}\mathbf{}\mathbf{6}\mathbf{}\mathbf{}\mathbf{u}$, calculate Qand compare it with the maximum energy of the emitted positron given above. (Hint:Let${\mathit{m}}_{\mathbf{C}}$ and${\mathit{m}}_{\mathbf{B}}$be the nuclear masses and then add in enough electrons to use the atomic masses.)

The radioactive decay,${}^{11}C\to {}^{11}B+e^{+}+v,T_{1/2}=20.3\min$

Maximum energy of the emitted positrons,$E_{max}=0.960\mathrm{MeV}$

Atomic mass of${}^{11}C,m_C=11.011434\mathrm{u}$,

Atomic mass of${}^{11}B,m_B=11.009305\mathrm{u}$ ,

Atomic mass of a positron,$m_e=0.0005486\mathrm{u}$

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.

Since, the positron has the same mass as an electron and the neutrino has negligible mass. Then, the energy considering its nuclear radii can be given as:

$∆m{c}^2=\left(m_B+m_e-m_C\right)c^2$

Now, since carbon has 6 electrons (see Appendix F and/or G) and boron has 5 electrons, we can add and subtract $6m_e$to the above expression and obtain the above expression as:

$$\begin{gathered} ∆m{c}^2=\left(m_B+7m_e-m_C-6m_e\right)c^2 \\ =\left(m_B+2m_e-m_C\right)c^2 \end{gathered}$$

The above final expression involves the atomic masses of the nuclidesas well an “extra” term corresponding to two electron masses.

Now, the above term considering the energy of an electron can be given as:

$∆m{c}^2=\left(m_C-m_B\right)c^2-2\left(0.511\mathrm{MeV}\right),\mathrm{as}{m}_e{c}^2=0.511\mathrm{MeV}......\left(1\right)$

Hence, the disintegration energy Q for this process is given by $Q=\left(m_C-m_B-2m_e\right)c^2$where, $m_C$and $m_B$are the atomic masses of ${}^{11}C$and ${}^{11}B$, respectively, and $m_e$is the mass of a positron.

Using the given data in equation (1), we can get the disintegration energy or the Q-value of the decay process as follows:

$$\begin{gathered} ∆m{c}^2=\left(11.011434\mathrm{u}-11.009305\mathrm{u}\right)\left(931.5\mathrm{MeV}/\mathrm{u}\right)-\left(1.022\mathrm{MeV}\right) \\ =0.961\mathrm{MeV} \end{gathered}$$

Hence, the value of Q is $0.961\mathrm{MeV}$and is almost the same value as given maximum energy of the positrons.