# How would you write a balanced nuclear equations for the emission of an alpha particle by polonium-209 and for the decay of calcium-45 to scandium-45?

Dec 5, 2015

${\text{_84^209"Po" -> ""_82^205"Pb" + }}_{2}^{4} \alpha$

${\text{_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + }}_{0}^{0} \nu$

#### Explanation:

The alpha decay of polonum-209, which consists of the emission of an alpha particle, will leave behind the atom of a different element.

Notice that the identity of the element changes after the decay. This means that you're dealing with a nuclear transmutation, a process that converts an atom of a given element into an atom of a different element.

Now, an alpha particle consists of

• two protons
• two neutrons

In essence, an alpha particle is a helium-4 nucleus, which means that you can see it as having anatomic number equal to $2$ and a mass number equal to $4$.

When a nucleus emits an alpha particle, its atomic number will decrease by $2$, since it's losing two protons, and its mass number will decrease by $4$, since it's losing two protons and two neutrons.

You can thus say that

${\text{_84^209"Po" -> ""_Z^A"X" + }}_{2}^{4} \alpha$

Balance the nuclear equation by balancing the atomic number and the mass number for both sides of the equation

$209 = A + 4 \text{ }$ and $\text{ } 84 = Z + 2$

The atomic number and the mass number of the resulting element will be

$A = 209 - 4 = 205 \text{ }$ and $\text{ } Z = 84 - 2 = 82$

A quick look in the periodic table will show that you're dealing with lead-205

${\text{_84^209"Po" -> ""_82^205"Pb" + }}_{2}^{4} \alpha$

Now for the decay of calcium-45 to scandium-45. Two important things to notice here

• you're once again dealing with a nuclear transmutation
• the mass number is conserved following the decay

Turn to the periodic table again and make a note of the atomic numbers of the two elements

${Z}_{C a} = 20 \text{ }$ and $\text{ } {Z}_{S c} = 21$

This means that you're looking for a nuclear decay that will increase the atomic number by $1$ and keep the mass number constant.

As you know, in beta decay, a neutron is being converted into a proton and a beta particle and an antineutrino are being emitted from the nucleus.

A beta particle is simply an electron $\beta = \text{_text(-1)^0"e}$

This means that you will have

${\text{_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + }}_{0}^{0} \nu$