Chapter U1.16, Problem 4E

Interpretation Introduction

Interpretation:

Nuclear equation for alpha decay of platinum - 191 is to be written.

Concept introduction:

Elements are found in one or more isotopic forms. Some isotopes have stable nuclei while some have unstable nuclei. Unstable nuclei undergo radioactive decay. An alpha particle is emitted from an unstable atom during radioactive decay.

Answer to Problem 4E

The nuclear equation for alpha decay of platinum-191 can be written as:

${}_{78}^{191}Pt{\to }_{76}^{187}Os{+}_2^{4}He$

Explanation of Solution

Given information:

An alpha particle is emitted from platinum - 191.

When an alpha particle is ejected, the process is called alpha decay. An alpha particle consists of 2 protons and 2 neutrons. It is same as the nucleus of a helium atom. Hence it can be represented by ${}_2^{4}He$. Nuclear reactions can be expressed as nuclear equations.

When an atom of platinum-191 undergoes alpha decay, an atom of new element is produced and an alpha particle is produced. New element has 2 protons and 2 neutrons less than platinum. From the periodic table one can find out the atomic number of platinum that is 78.

So,

Atomic number of new element is 76.

The mass number of new element is 191-4=187

From the periodic table one can find out the element with atomic number of 76. It is osmium.

The nuclear equation can be written as:

${}_{78}^{191}Pt{\to }_{76}^{187}Os{+}_2^{4}He$

Conclusion

Alpha decay of platinum - 191 results in formation of osmium - 187 and a helium atom (alpha particle).