Test Index

WAEC Physics 2015 Paper

Question : 49

Total: 50

^{226}_{88}Ra

Solution:

Concept:

Each alpha particle is a helium nucleus

\left(^{4}_{2}\mathrm{He}\right)

It reduces the mass number by 4 and the atomic number by 2.

Explanation:

Initial nuclide:

^{226}_{88}\mathrm{Ra}

After the first alpha emission:

^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + ^{4}_{2}\mathrm{He}

After the second alpha emission:

^{222}_{86}\mathrm{Rn} \rightarrow ^{218}_{84}\mathrm{Po} + ^{4}_{2}\mathrm{He}

Thus, the polonium nuclide formed is

^{218}_{84}\mathrm{Po}

Answer:

^{218}_{84}\mathrm{Po}

Hence, the correct option is D.

Go to Question:

Prev Question

Next Question