Question

how is the nucleus changed by the decay below?
226 88ra→4 2he+222 86rn
a positron and a beta particle annihilate each other
the mass decreases by 4 and the atomic number
decreases by 2
a proton is converted to a neutron
a neutron is converted to a proton

Explanation:

Step1: Analyze the mass number

The mass number of \(^{226}_{88}Ra\) is \(226\), and the sum of the mass numbers of \(^{4}_{2}He\) and \(^{222}_{86}Rn\) is \(4 + 222=226\). The mass number of the nucleus (parent \(^{226}_{88}Ra\)) decreases by \(4\) (as \(226-222 = 4\)) due to the emission of \(^{4}_{2}He\) (alpha - particle).

Step2: Analyze the atomic number

The atomic number of \(^{226}_{88}Ra\) is \(88\), and the sum of the atomic numbers of \(^{4}_{2}He\) and \(^{222}_{86}Rn\) is \(2+86 = 88\). The atomic number of the nucleus (parent \(^{226}_{88}Ra\)) decreases by \(2\) (as \(88 - 86=2\)) due to the emission of \(^{4}_{2}He\) (alpha - particle).

Answer:

the mass decreases by 4 and the atomic number decreases by 2