Q-40PE - College Physics (Urone)

Question

a) Write the complete \({{\rm{\beta }}^{\rm{ + }}}\) decay equation for \({}^{{\rm{11}}}{\rm{C}}\).(b) Calculate the energy released in the decay. The masses of \({}^{{\rm{11}}}{\rm{C}}\) and\({}^{{\rm{11}}}{\rm{B}}\) are \(11.011433\) and\(11.009305\,{\rm{u}}\) , respectively.

Step-by-Step Solution

Answer

(a) The equation is\({}_6^{11}{C_5} \to {}_5^{11}{B_6} + {\beta ^ + } + {v_e}\).

(b) The energy released is\(0.958\,{\rm{MeV}}\).

Define energy

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

 

Explanation

(a) The equation for \({{\rm{\beta }}^{\rm{ + }}}\) is\({}_Z^A{X_N} \to {}_{Z - 1}^A{Y_{N + 1}} + {\beta ^ + } + {v_e}\).

The electron's neutrino is \({{\rm{v}}_{\rm{e}}}\). 

\(A = Z + N\) is also known. As a result, the decay equation is

\({}_6^{11}{C_5} \to {}_5^{11}{B_6} + {\beta ^ + } + {v_e}\).

Evaluating the energy

(b) Use the following relationship to calculate the change in mass that leads to the released energy as a result of the decay,

\(\begin{align}{\underline{\phantom{xx}}}\Delta m &= {m_C} - {m_B} - 2{m_\beta }\\ &= 11.011433\,{\rm{u}} - 11.009305\,{\rm{u}} - 2 \times 0.00055\,{\rm{u}}\\ &= 1.028 \times {10^{ - 3}}\,{\rm{u}}\end{align}\)

Einstein's mass/energy equation is as follows,

\(\begin{align}{\underline{\phantom{xx}}}E &= \Delta m{c^2}\\ &= 1.028 \times {10^{ - 3}}\,{\rm{u}} \times \left( {931.5\,{\rm{MeV/u}}{{\rm{c}}^{\rm{2}}}} \right) \times {c^2}\\ &= 0.958\,{\rm{MeV}}\end{align}\)

Therefore, the energy is \(0.958\,{\rm{MeV}}\).