Question

3 what is the balanced nuclear equation for the alpha decay of radium - 226 into radon?
a ( _{88}^{226}ra
ightarrow_{86}^{222}rn + _{2}^{4}he )
b ( _{88}^{226}ra
ightarrow_{86}^{222}rn + _{-2}^{0}e )
c ( _{88}^{226}ra
ightarrow_{86}^{222}rn + 2(_{1}^{2}he) )
d ( _{88}^{226}ra
ightarrow_{86}^{222}rn + 2(_{-1}^{0}e) )

Explanation:

Step1: Recall alpha decay

In alpha decay, an atom emits an alpha particle (\(^{4}_{2}\text{He}\)).

Step2: Check mass and atomic numbers

For radium - 226 (\(^{226}_{88}\text{Ra}\)):

• Mass number: \(226\)
• Atomic number: \(88\)

After emitting an alpha particle (\(^{4}_{2}\text{He}\)):

• New mass number: \(226 - 4=222\)
• New atomic number: \(88 - 2 = 86\)

Which corresponds to radon (\(^{222}_{86}\text{Rn}\))

Answer:

A. \(^{226}_{88}\text{Ra}\to^{222}_{86}\text{Rn}+^{4}_{2}\text{He}\)