# What are the atomic number and atomic mass of the element formed when 218Po emits a beta particle? What are they if the polonium emits an alpha particle?

Jul 25, 2018

Alpha decay

$\text{_84^218"Po}$$\rightarrow$$\text{_82^214"Pb" + ""_2^4"He}$

Beta decay

$\text{_84^218"Po}$$\rightarrow$$\text{_85^218"At} + {e}^{-} + {\overline{\nu}}_{e}$

#### Explanation:

$\text{_84^218"Po}$ undergoes alpha decay and beta minus decay.
https://en.wikipedia.org/wiki/Isotopes_of_polonium

Alpha Decay

In alpha decay, an unstable nucleus will emit an alpha particle, which is a helium nucleus composed of two protons and neutrons. The mass number of the daughter isotope will be reduced by four, and the atomic number will be reduced by 2. The alpha decay of $\text{_84^218"Po}$ produces $\text{_82^214"Pb}$.

$\text{_84^218"Po}$$\rightarrow$$\text{_82^214"Pb" + ""_2^4"He}$

Beta Minus Decay.

In beta minus decay, a neutron in an unstable nucleus decays into a proton, an electron, and an antineutrino: $\text{n"^0}$$\rightarrow$${\text{p}}^{+} + {e}^{-} + {\overline{\nu}}_{e}$ The daughter isotope will have the same mass number, but the atomic number increases by one.

The beta minus decay of $\text{_84^218"Po}$ produces $\text{_85^218"At}$.

$\text{_84^218"Po}$$\rightarrow$$\text{_85^218"At} + {e}^{-} + {\overline{\nu}}_{e}$

Jul 25, 2018
• $A ' = 218$ and $Z ' = 85$ when ${\textcolor{w h i t e}{l}}_{84}^{218} \text{Po}$ undergoes beta minus decay.
• $A ' ' = 214$ and $Z ' ' = 82$ when ${\textcolor{w h i t e}{l}}_{84}^{218} \text{Po}$ undergoes alpha decay.

#### Explanation:

An atom releases an electron (${e}^{-}$) as it undergoes beta minus decay. A neutron (a subatomic particle of $A = 1$, $Z = 0$) is converted to a proton ($A = 1$, $Z = 1$) in this process:

${\textcolor{w h i t e}{l}}_{0}^{1} {n}^{0} \to {\textcolor{w h i t e}{l}}_{1}^{1} {p}^{+} + {e}^{-}$

Thus for each electron (beta minus particle) emitted, the atomic number of the atom increase by one whereas its mass number stays the same. Therefore one would expect the beta minus decay product of ${\textcolor{w h i t e}{l}}_{84}^{218} \text{Po}$ to be of

• mass number $A ' = A = 218$ and
• atomic number $Z ' = Z + 1 = 85$

An atom emits a bare helium nucleus ("alpha particle" ${\textcolor{w h i t e}{l}}_{2}^{4} \text{He}$ for which $A = 4$ and $Z = 2$) as it undergoes alpha decay.

The alpha particle contains two protons bonded to two neutrons, all of the four nucleons are directly removed from the parent nucleus (${\textcolor{w h i t e}{l}}_{84}^{218} \text{Po}$ in this case.) Thus the atomic number $\text{Z}$ of the daughter nucleus of the alpha decay shall be smaller than that of the parent nucleus by $2$ whereas the mass number $\text{A}$ smaller by $2 + 2 = 4$. Therefore the daughter nuclei produced in the alpha decay of ${\textcolor{w h i t e}{l}}_{84}^{218} \text{Po}$ would have

• mass number $A ' ' = A - 4 = 214$
• atomic number $Z ' ' = Z - 2 = 82$