Answer
Nuclear equation for the alpha decay of each nuclide is as follows:
Work Step by Step
When one alpha ray is emitted from a radioactive element, atomic number of that element i.e. no. of protons are decreased by 2 unit and mass number of the element is decreased by 4 unit. It is because two protons and two neutrons are emitted from the nucleus of that nuclide in the form of helium as alpha rays.
Now, see every nuclear equation above.
After alpha decay, Po-218 is converted to Pb-214 and one alpha ray is emitted in the form of He-4 atom.
Similarly, Po-214 is converted to Pb-210 after alpha decay.
Po-210 is converted to Pb-206 after alpha emission and in the end Th-227 is converted to Ra-223 after alpha decay.
When writing a nuclear equation each element is represented in symbol. (Process to write element in the form of symbol: Write the chemical symbol of the element, write the atomic number of the nuclide at left subscript position of chemical symbol and write mass number at the left superscript position of chemical symbol for a particular element). Alpha emission is denoted by writing Helium atom in the form of symbol on the right side of the equation.
After writing the nuclear equation, the total mass number (sum of mass number for each nuclide on the left side) on the left side will be equal to the total mass number on the right side (sum of mass number for each nuclide on the right side). Similarly, total atomic number on the left side will be equal to the total atomic number on the right side.
General form of nuclear equation after alpha decay is as follows:
