Question

Give equation for radium-222 when it undergoes decay by emission of an α-particle.

\(^{226}_{88}Ra\) \(\longrightarrow\) \(^{222}_{86}X\) + \(^{4}_{2}He\)

Answer 1

 \(^{226}_{88}Ra\) \(\longrightarrow\) \(^{222}_{86}X\) + \(^{4}_{2}He\)

Thus, atomic number of element ‘X’ will be 86 and atomic mass number will be 222.