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
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:
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.