Question

Write symbolically the nuclear β⁺ decay process of ${\displaystyle {}_6^{11}C}$ Is the decayed product X an isotope or isobar of (${\displaystyle {}_6^{11}C}$)? Given the mass values m (${\displaystyle {}_6^{11}C}$) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

Answer 1

In β⁺ decay, the atomic number Z of the nucleus goes up by 1.

The nuclear β⁺ decay process of ${\displaystyle {}_6^{11}C}$ is ${\displaystyle {}_6^{11}C\to {}_5^{11}B+e^{+}+v}$

Given :

${\displaystyle \left({}_6^{11}C\right)}$= 11.011434 u

m (X) =${\displaystyle \left({}_5^{11}B\right)}$= 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 ${\displaystyle {}_6^{11}C}$ is given as:

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

If atomic masses are used instead of nuclear masses, then we have to add 6m_e in the case of ¹¹C and 5m_e in the case of ¹¹B

Hence, eq.(1) reduces to

${\displaystyle Q=[m\left({}_6^{11}C\right)-m\left({}_5^{11}B\right)-2m_e]C^2}$

Q= [11.011434-11.009305-2x 0.000548]C²

Q = 0.001033 C² u

1 u = 931.5 Mev/C²

Q = 0.001033 x 931.5 = 0.962 Mev