# What is the daughter nucleus produced when 103 Mo undergoes beta decay?

Apr 2, 2016

""_43^103Tc or ""_41^103Nb depending on the type of $\beta$-decay.

#### Explanation:

$\beta$-decay is when a neutron is transformed into a proton, releasing an electron, or a proton becomes a neutron, releasing a positron.

Mo-103 is a molybdenum isotope with a mass number of 103 and (from the periodic table) an atomic number of 42. We write this as ""_42^103Mo.

Lets assume that a neutron is turning into a proton an emitting an electron. In this case, the mass number stays the same, because neutrons and protons have roughly the same mass, though the atomic number increases by one, meaning it is in fact a new element.

Looking on the periodic table for element no. 43, we find it is technetium, Tc, ""_43^103Tc.

In the form of an equation, this would look like

${\text{_ 42^103Mo -> ""_ 43^103Tc + ""_"-1}}^{0} \beta$

You can check this equation by adding up the top and bottom numbers next to the chemical symbols.

$42 = 43 + \left(- 1\right)$
$103 = 103 + 0$

If the maths is correct, it is a plausible decay scenario.

Positron decay looks similar. A positron is essentially a positive electron.

${\text{_42^103Mo -> ""_41^103Nb + }}_{1}^{0} \beta$