Chapter 30, Problem 34P

To determine

The maximum amount of energy of ${\text{β}}^{\text{+}}$ particles released when ${}_6{}^{11}\text{C}$ decays to ${}_5{}^{11}\text{B}$. The maximum energy the neutrino can have. The minimum energy of each.

Answer to Problem 34P

Solution:

${\text{KE}}_{\text{max}}=0.9612\text{MeV}$. ${\text{KE}}_{\text{min}}=0\text{MeV}$

Explanation of Solution

Formula used:

Binding energy is $\Delta m{c}^2$.

Solution:

The decay is ${}_6{}^{11}\text{C}\to {}_5{}^{11}\text{B}+{\text{β}}^{\text{+}}+\nu$. The sum of the kinetic energy of the β+ particle and the energy of the neutrino must be from the mass difference, so the kinetic energy of the neutrino will range from $0.9612\text{MeV}$ to 0 MeV.

Calculation:

Mass of neutron ${}_6{}^{11}\text{C}$ is $11.011434\text{u}$. Mass of ${}_5{}^{11}\text{B}$ is $11.009305\text{u}$. Mass of electron is $0.0005486\text{u}$. Mass defect is $\Delta m=m\left({}_6{}^{11}\text{C}\right)-m\left({}_5{}^{11}\text{B}\right)-2m_e=11.011434\text{u}-11.009305\text{u }-2\times 0.0005486\text{u}=\text{0}\text{.0010318}\text{u}$

Energy released is $\text{0}\text{.0010318}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=0.9612\text{MeV}$

Conclusion:

Mass is equivalent to energy. Mass in a decay reaction cannot increase. The change in mass appears as energy.There are three types of neutrinos: electron, muon and tau. Here we take the electron as neutrino. ${\text{KE}}_{\text{max}}=0.9612\text{MeV}$. ${\text{KE}}_{\text{min}}=0\text{MeV}$