Question

Write symbolically the nuclear${\beta }^{+}$ decay process of $6^{11}\text{C}$ . Is the decayed product X an isotope or isobar of $\left(6^{11}\text{C}\right)$? Given the mass values m $\left(6^{11}\text{C}\right)$ = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

Answer 1

In ${\beta }^{+}$ decay, the atomic number Z of the nucleus goes up by 1

The nuclear ${\beta }^{+}$ decay process of $6^{11}\text{C}$ is

${}_6^{11}\text{C}{\to }_5^{11}\text{B+}{e}^{+}+v$

Given
${}_6^{11}\text{C}11.011434u$
$m\left(x\right){=}_5^{11}\text{B}=11.009305$u

Mass of an electron or positron = 0.000548 u

c = speed of light

The Q value of the nuclear masses of the ${}_6^{11}\text{C}$ is given as:

$Q=\left[m{(}_6^{11}\text{C}\right)-m\left(m\right(X\left)\right)+m_e]C^2$.....(1)

If atomic masses are used instead of nuclear masses then we have to add$6m_e$ in the case of ${}^{11}C$ and 5m in the case of ${}^{11}B$ Hence, eq.(1) reduces to

$Q=\left[m{(}_6^{11}\text{C}\right)-m{(}_5^{11}\text{B})-2m_e]C^{2}Q=[11.011434-11.009305-2\times 0.000548]C^{2}Q=0.001033C^{2}u1u=931.5Mev/C^{2}Q=0.001033\times 931.5=0.962Mev$