Question

if an atom of polonium-218 were to undergo alpha emission, followed by two beta emissions, which nuclide would result? lead-214 bismuth-214 polonium-214 lead-210

Answer 1

Po, (218, 84) - Alpha Particle (4, 2) = (218 - 4, 84 - 2) = (214 , 82)

(214 , 82) - 2Beta = (214, 82 - 2*(-1)) = (214 , 84)

Polonium has atomic number of 84.

So answer = Polonium-214

Answer 2

The resulting nuclide will be polonium-214.

Step-by-step explanation:

Alpha decay: It is a process in which alpha particles is emitted when a heavier nucleus decays into lighter nuclei. The alpha particle released has a charge of +2 units.

The general alpha decay reaction is given as:

`_Z^A\textrm{X}\rightarrow _(Z-2)^(A-4)\textrm{Y}+_2^4\alpha`

Beta-decay: It is a process in which a neutron gets converted into a proton and an electron releasing as a beta-particle. The beta particle released carries a charge of -1 units.

`_Z^A\textrm{X}\rightarrow _(Z+1)^A\textrm{Y}+_(-1)^0\beta`

Atom of polonium-218 after alpha decay and two emissions:

1)
`_(84)^(218)\textrm{Po}\rightarrow _(82)^(214)\textrm{Pb}+_2^4\alpha`

2)
`_(82)^(214)\textrm{Pb}\rightarrow _(83)^(214)\textrm{Bi}+_(-1)^0\beta`

3)
`_(83)^(214)\textrm{Bi}\rightarrow _(84)^(214)\textrm{Po}+_(-1)^0\beta`

The resulting nuclide will be polonium-214.