Question

Question: The radioactive nuclide ${}_{\mathbf{29}}{}^{\mathbf{64}}\mathbf{Cu}$decays by beta emission to ${}_{\mathbf{30}}{}^{\mathbf{64}}\mathbf{Zn}$or by positron emission to ${}_{\mathbf{28}}{}^{\mathbf{64}}\mathbf{Ni}$. The maximum kinetic energy of the beta particles is 0.58 MeV, and that of the positrons is 0.65 MeV. The mass of the neutral atom is 63.92976 u.

1. Calculate the mass, in atomic mass units, of the neutral atom ${}_{\mathbf{30}}{}^{\mathbf{64}}\mathbf{Zn}$atom.
2. Calculate the mass, in atomic mass units, of the neutral atom ${}_{\mathbf{28}}{}^{\mathbf{64}}\mathbf{Ni}$atom.

Short answer

1. The atomic mass of ${}_{30}{}^{64}\mathrm{Zn}$is$63.9282\mathrm{u}$.
2. The atomic mass of ${}_{28}{}^{64}\mathrm{Ni}$is $63.9265\mathrm{u}$.

Step-by-step solution

(a) Atomic mass of  Zn3064 atom

The kinetic energy of the Beta particle$=0.58\mathrm{MeV}$

We can calculate the mass difference from the kinetic energy of the beta particle.

$\begin{array}{rcl}∆\mathrm{E}& =& {\mathrm{c}}^2∆\mathrm{m}\\ ∆\mathrm{m}& =& \frac{∆\mathrm{E}}{{\mathrm{c}}^2}\\ ∆\mathrm{m}& =& \frac{0.58}{931.5}\mathrm{u}\\ & =& 0.000624\end{array}$

From the mass difference, we can calculate the atomic mass of ${}_{30}{}^{64}\mathrm{Zn}$.

$$\begin{gathered} ∆\mathrm{m}=\mathrm{m}\left({}_{30}{}^{64}\mathrm{Zn}\right)-\mathrm{m}\left({}_{29}{}^{64}\mathrm{Cu}\right) \\ \Rightarrow 0.000624\mathrm{u}+\mathrm{m}\left({}_{29}{}^{64}\mathrm{Cu}\right)=\mathrm{m}\left({}_{30}{}^{64}\mathrm{Zn}\right) \\ \Rightarrow 0.000624+63.9276\mathrm{u}=\mathrm{m}\left({}_{30}{}^{64}\mathrm{Zn}\right) \\ \Rightarrow \mathrm{m}\left({}_{30}{}^{64}\mathrm{Zn}\right)=63.9282\mathrm{u} \end{gathered}$$

 Step 2: Atomic mass of the Ni2864 atom

The kinetic energy of the positron particle$=0.65\mathrm{MeV}$

We can calculate the mass difference from the kinetic energy of the positron particle.

$\begin{array}{rcl}∆\mathrm{E}& =& {\mathrm{c}}^2∆\mathrm{m}\\ ∆\mathrm{m}& =& \frac{∆\mathrm{E}}{{\mathrm{c}}^2}\\ ∆\mathrm{m}& =& \frac{0.65}{931.5}\mathrm{u}\\ & =& 0.0006977\end{array}$

From the mass difference, we can calculate the atomic mass of${}_{28}{}^{64}\mathrm{Ni}$.
role="math" localid="1661277114735" $$\begin{gathered} ∆\mathrm{m}=\mathrm{m}\left({}_{28}{}^{64}\mathrm{Ni}\right)+2\mathrm{m}\left({}_1{}^0\mathrm{e}^{+}\right)-\mathrm{m}\left({}_{29}{}^{64}\mathrm{Cu}\right) \\ \Rightarrow 0.000624\mathrm{u}+\mathrm{m}\left({}_{29}{}^{64}\mathrm{Cu}\right)-2\mathrm{m}\left({}_1{}^0\mathrm{e}^{+}\right)=\mathrm{m}\left({}_{28}{}^{64}\mathrm{Ni}\right) \\ \Rightarrow 0.000624+63.9276\mathrm{u}-2\times 0.0005485\mathrm{u}=\mathrm{m}\left({}_{28}{}^{64}\mathrm{Ni}\right) \\ \Rightarrow \mathrm{m}\left({}_{28}{}^{64}\mathrm{Ni}\right)=63.9265\mathrm{u} \end{gathered}$$