In the reaction that produces 123I, is there a minimum kinetic energy the protons need to make the reaction go? (a) No, because the proton has a smaller mass than the neutron. (b) No, because the total initial mass is smaller than the total final mass. (c) Yes, because the proton has a smaller mass than the neutron. (d) Yes, because the total initial mass is smaller than the total final mass.

We are given the following atomic masses:

For ${}^{123}Te$isrole="math" localid="1667818806552" $M\left({}^{123}\mathrm{Te}\right)=122.90427u$

For ${}^{123}l$is$M\left({}^{123}\mathrm{I}\right)=122.905589u$

For ${}^{1}H$is$M\left({}^1\mathrm{He}\right)=1.007825u$

And the mass of the neutron is $m_n=1.008665$.

${}_{52}{}^{123}\mathrm{Te}+p\to {}_{53}{}^{123}\mathrm{l}+n$

In order to know if the proton needs an initial kinetic energy or not, we need to calculate the Q-value of the reaction which is given by:

$Q=(M_A+M_B-M_C-M_D)\times 931.5MeV/u$ (1)

Where $M_A$and $M_B$are the masses of the reacting nuclei and $M_C$and $M_D$are the masses of the resulting nuclei.

The atomic masses involve the masses of the orbiting electrons, but if we replace the mass of proton by the mass of hydrogen atom, the number of electrons on th both sides is equal and cancel each other in our calculation.

so, we substitute our value for the given mass into equation (0 so we get the Q

$$\begin{gathered} Q=\left[M{(}^{123}Te\right)+M{(}^{1}H)-M{(}^{123}I)-m_n]\times 931.5MeV/u \\ =[122.904270\mathrm{u}+1.007825\mathrm{u}-122.905589\mathrm{u}-1.008665\mathrm{u}]\times 931.5\mathrm{MeV}/\mathrm{u} \\ =[-0.002159\mathrm{u}]\times 931.5\mathrm{MeV}/\mathrm{u} \\ =-2.011\mathrm{MeV} \end{gathered}$$

Since the Q-value is negative, the reaction is endoergic.

Thus, there is minimum kinetic energy the protons need to make the reaction go.

Therefore, the right choice is (d) Yes, because the total initial mass is smaller than the total final mass.