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

Answer:

The resulting nuclide will be polonium-214.

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.