Answer
See answers.
Work Step by Step
a. Consider the mass and charge numbers. When $^{32}_{15}P$ emits an electron, its mass number stays the same and the atomic number increases by 1. Therefore the daughter nucleus is $^{32}_{16}S$.
b. Neglect the recoil energy of the sulfur atom, and the neutrino’s energy. The energy released is the maximum kinetic energy of the electron.
$$E_{release}=KE=\left( m(^{32}_{15}P) -m(^{32}_{16}S) \right)c^2$$
$$ m(^{32}_{16}S ) =m(^{32}_{15}P)-\frac{KE}{c^2}$$
$$ =31.973908u-\left(\frac{1.71MeV}{c^2} \frac{u}{931.49MeV/c^2}\right)$$
$$ m(^{32}_{16}S ) =31.97207u$$
