# Question #d3263

Oct 7, 2017

Actually, germanium-71 does not undergo beta decay.

#### Explanation:

The interesting thing to point out here is that germanium-71 does not undergo beta decay, it actually undergoes electron capture $\to$ see here.

When germanium-71 undergoes electron capture, its nucleus captures an electron from one of its inner energy shells, which results in the conversion of a proton to a neutron.

Consequently, the atomic number of the nuclide will decrease by $1$. On the other hand, its mass number will remain unchanged.

At this point, the nucleus emits an electron neutrino, ${\nu}_{e}$.

The balanced nuclear equation that describes the electron capture of germanium-71 looks like this

$\text{_ 32^71"Ge" + ""_(-1)^(color(white)(-)0)"e" -> ""_31^71"Ga} + {\nu}_{e}$

Notice that you have

• $71 + 0 = 71 \to$ conservation of mass

• $32 + \left(- 1\right) = 31 \to$ conservation of charge

The resulting nuclide is galium-71.

Now, in a beta decay, the nucleus of a radioactive nuclide emits a beta particle, which is another name given to an electron, and an electron antineutrino, ${\overline{\nu}}_{e}$, as a result of the fact that a neutron is being converted into a proton.

This time, the atomic number of the nuclide increases by $1$ and its mass number remains unchanged.

So assuming that germanium-71 undergoes beta decay, the balanced nuclear equation looks like this

$\text{_ 32^71"Ge" -> ""_ 33^71"As" + ""_(-1)^(color(white)(-)0)"e} + {\overline{\nu}}_{e}$

Once again, you have

• $71 = 71 + 0 \to$conservation of mass
• $32 = 33 + \left(- 1\right) \to$ conservation of charge

This time, the resulting nuclide would be arsenic-33.