Chapter 19, Problem 19.1P

Interpretation Introduction

Interpretation:

The nuclear equation for the alpha decay of Po- $216$ is to be stated.

Concept introduction:In the alpha decay, mass number of the nuclide decreases by $4$ units and atomic number decreases by $2$ units. Alpha decay occurs through emission of a helium nucleus ( ${}_2^4\text{He}$ ).

To determine:A nuclear equation corresponding to the alpha decay of Po- $216$ .

Answer to Problem 19.1P

Solution:

The nuclear equation is,

${}_{84}^{216}\text{Po}{\to }_{82}^{212}\text{Pb}+{}_2^4\text{He}$

Explanation of Solution

The given nuclide is Po- $216$ . Atomic number of polonium is $84$ . In alpha decay, mass number of the nuclide decreases by $4$ units. Therefore, mass number of daughter nuclide will be $216-4=212$ .

In the alpha decay, atomic number of nuclide decreases by $2$ units. Therefore, atomic number of the daughter nuclide will be $84-2=82$ . According to periodic table, atomic number of lead (Pb) is $82$ . Hence, the daughter nuclide is ${}_{82}^{212}\text{Pb}$ .

Alpha decay occurs through emission of a helium nucleus ( ${}_2^4\text{He}$ ).The nuclear equation is,

${}_{84}^{216}\text{Po}{\to }_{82}^{212}\text{Pb}+{}_2^4\text{He}$

Conclusion

The nuclear equation is,

${}_{84}^{216}\text{Po}{\to }_{82}^{212}\text{Pb}+{}_2^4\text{He}$