Question:

The radionuclide 11C decays according to

${ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v: \quad \mathrm{T}_{1 / 2}=20.3 \mathrm{~min}$

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

$m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}$ and $m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}$,

calculate and compare it with the maximum energy of the positron emitted

Solution:

The given nuclear reaction is:

${ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v$

Half life of ${ }_{6}^{11} \mathrm{C}$ nuclei, $T_{1 / 2}=20.3 \mathrm{~min}$

Atomic mass of $m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}$

Atomic mass of $m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}$

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the $Q$-value $(\Delta Q)$ of the nuclear masses of the ${ }_{6}^{11} \mathrm{C}$ nucleus is given as:

$\Delta Q=\left[m^{\prime}\left({ }_{6} \mathrm{C}^{11}\right)-\left[m^{\prime}\left({ }_{5}^{11} \mathrm{~B}\right)+m_{e}\right]\right] c^{2}$     ....(1)

Where,

me = Mass of an electron or positron = 0.000548 u

= Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add $6 m_{e}$ in the case of ${ }^{11} \mathrm{C}$ and $5 m_{e}$ in the case of ${ }^{11} \mathrm{~B}$.

Hence, equation (1) reduces to:

$\Delta Q=\left[m\left({ }_{6} \mathrm{C}^{11}\right)-m\left({ }_{5}^{11} \mathrm{~B}\right)-2 m_{e}\right] c^{2}$

Here, $m\left({ }_{6} \mathrm{C}^{11}\right)$ and $m\left({ }_{5}^{11} \mathrm{~B}\right)$ are the atomic masses.

∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2

= (0.001033 c2) u

But 1 u = 931.5 Mev/c2

∴ΔQ = 0.001033 × 931.5  0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.