Question

${}^{11}C$ undergoes ${\beta }^{+}$ decay to ${}^{11}B.$ What is the maximum energy a positron emitted in the process can possess?
Atomic masses: ${}^{11}C:11.0114u;{}^{11}B:11.0093u;e^{-}:\mathrm{0.005486.}{c}^2=931\frac{MeV}{u}$.

• A. Can't be uniquely determined.
• B. 1.44 MeV
• C. 833.6 keV
• D. 933.6 keV

Answer 1

The correct option is D 933.6 keV

Let us consider a general ${\beta }^{+}$ decay process, as you have learned:
${}_Z^{A}X{\to }_{Z-1}^{A}Y+e^{+}+v.$
Now, mass of the nucleus, ${}_Z^{A}X$, is equal to the mass of the atom - mass of the electrons
$=m{(}_Z^{A}X)-Z{m}_e.$
Here m${(}_Z^{A}X)$ is the atomic mass of ${}_Z^{A}X$, Z is the atomic number (which is equal to the number of electrons) and $m_e$ is the mass of an electron.
Similarly, mass of nucleus ${}_{Z-1}^{A}Y=m{(}_{Z-1}^{A}Y)-(Z-1)m_e.$
Now, mass of positron $\left(e^{+}\right)$ = mass of electron = $m_e.$
And, the neutrino, ν, is like a photon in that it has zero rest mass!
So, the total mass of the products = mass of nucleus
${}_{Z-1}^{A}Y$ + mass of $e^{+}$
$=m{(}_{Z-1}^{A}Y)-(Z-1)m_e+m_e$
$=m{(}_{Z-1}^{A}Y)-(Z+2)m_e.$
Therefore, the mass lost = mass of parent nuclei - total mass of products
$=m{(}_Z^{A}X)-Z{m}_e-m{(}_{Z-1}^{A}Y)-(Z+2)m_e$
$=m{(}_Z^{A}X)-m{(}_{Z-1}^{A}Y)-2m_e.$
And the equivalent energy associated with this lost mass, or the Q-Value
$(=m{(}_Z^{A}X)-m{(}_{Z-1}^{A}Y)-2m_e)c^2.$
Here the parent and daughter nuclei are ${}^{11}Cand{}^{11}B$
respectively. So, the $Q-Value=(11.0114u-11.0093u-2\times 0.005486u)\times 931\frac{MeV}{u}$.
=0.933607 MeV=933.6 keV.
Now, this energy is shared by the neutrino and positron, in general, in some ratio. Therefore, the maximum possible energy the positron can have is the total energy! This can happen when none of the energy is taken away by the neutrino. Hence, here, the maximum energy a positron can possess is 933.6 keV.