Question

How do you write a balanced nuclear equation for alpha decay of Po-$218$?

Answer 1

While writing a balanced nuclear equation, we should follow the law of conservation of charge and law of conservation of mass. We should know that during an alpha decay, an alpha particle is emitted. Alpha particles consist of two protons and two neutrons bound together into a particle identical to a helium-$4$ nucleus. This means that the atomic number of an alpha particle is two and the mass number of an alpha particle is four.

Complete step by step solution:
We know that during an alpha decay, the radioactive nuclide (in our case Po-$218$) emits an alpha particle ${^2_4}\alpha$ which is essentially the nucleus of a helium- $4$ atom.
We should remember these points while writing a nuclear equation for an alpha decay,
The mass number of the resulting nuclide will be less than four as compared to the parent radioactive nuclide. This is because the alpha particle has a mass number of four due to the presence of two protons and two neutrons. Law of conservation of mass is fulfilled now.
The atomic number of the resulting nuclide will be less than two as compared to the parent radioactive nuclide. This is because the alpha particle has an atomic number of two due to the presence of two protons. Law of conservation of charge is fulfilled now.
Now the nuclear equation for alpha decay of Po-$218$ can be written as:
$_{84}^{218}Po \to _{84 - 2}^{218 - 4}X + _2^4\alpha _{}^{}$
$_{84}^{218}Po \to _{82}^{214}X + _2^4\alpha$
The element with atomic number $82$ is Pb (Lead).
Hence, the balanced nuclear equation for alpha decay of Po-$218$ is;
$_{84}^{218}Po \to _{82}^{214}Pb + _2^4\alpha$
Which is the required answer.

Note: It should be noted that we should be quite familiar with the periodic table while solving these types of questions on radioactivity. You should also study other reactions on radioactivity too. Law of conservation of mass and law of conservation of charge has to be satisfied in every reaction of radioactivity.