Question

Complete the following decay equation:

_{86}^{226}\textrm{Ra}\rightarrow_{Z}^{A}+_{2}^{4}\textrm{He}

Mass number 226 = A + 4

A = 226 - 4 = 222

Atomic number 86 = Z + 2

Z = 86 - 2 = 84

_{86}^{226}\textrm{Ra}\rightarrow_{84}^{222}+_{2}^{4}\textrm{He}

Question

Polonium-210 decays to lead-206. Polonium (Po) has atomic number 84 and lead (Pb) has atomic number 82.

Which type of decay occurs?

_{84}^{210}\textrm{Po}\rightarrow_{82}^{206}\textrm{Pb}+_{Z}^{A}\textrm{X}

Mass number 210 = 206 + A

A = 210 - 206 = 4

Atomic number 84 = 82 + Z

Z = 84 - 82 = 2

X has mass number 4 and atomic number 2 and so is an alpha particle _{2}^{4}\textrm{He}

_{84}^{210}\textrm{Po}\rightarrow_{82}^{206}\textrm{Pb}+_{2}^{4}\textrm{He}