Chapter 20, Problem 1SAQ

Interpretation Introduction

Interpretation: The daughter nuclei formed by the alpha decay of polonium-214 is to be identified from the given options.

a) ${\mathrm{\hspace{0.17em}}}_{86}^{218}\text{Rn}$

b) ${\mathrm{\hspace{0.17em}}}_{85}^{214}\text{At}$

c) ${\mathrm{\hspace{0.17em}}}_{83}^{214}\text{Bi}$

d) ${\mathrm{\hspace{0.17em}}}_{82}^{210}\text{Pb}$

Answer to Problem 1SAQ

Solution: Correct option is (d).

Explanation of Solution

The original atom is called parent nuclide, and the product of the decay is called the daughter nuclide. The sum of the atomic numbers on both sides of a nuclear equation must be equal. Also, the sum of the mass numbers on both sides must also be equal.

The mass number decreases by 4 and the atomic number decreases by 2 during alpha decay. Referring to the periodic table, the atomic number of polonium (Po) is 84.Start with the symbol for $Po\text{-}214$ on the left side of the equation and the symbol for an alpha particle $\left({}_2^{4}He\right)$ on the right side.

${}_{84}^{214}Po\to {}_{?}^{?}?+{}_2^{4}He$

Equalize the sum of the mass numbers and the sum of the atomic numbers on both sides of the equation by writing the appropriate mass number and the atomic number for the unknown daughter nuclide.

${}_{84}^{214}Po\to {}_{82}^{210}?+{}_2^{4}He$

Refer to the periodic table to deduce the identity of the unknown daughter nuclide from its atomic number and write its symbol. Since the atomic number is 82, the daughter nuclide will be lead (Pb).

${}_{84}^{214}Po\to {}_{82}^{210}Pb+{}_2^{4}He$

Thus, the daughter nuclide will be lead (${}_{82}^{210}Pb$).

Conclusion

The daughter nuclei formed by the alpha decay of polonium-214 is ${\mathrm{\hspace{0.17em}}}_{82}^{210}\text{Rn}$.