When the nuclide polonium-218 undergoes alpha decay:
The name of the product nuclide is. _
The symbol for the product nuclide is. _
Write a balanced nuclear equation for the following:
The nuclide bismuth-210 undergoes alpha emission.
(Use the lowest possible coefficients.)
The correct answer and explanation is :
Answer:
- The name of the product nuclide: Lead-214
- The symbol for the product nuclide: (\text{Pb-214})
Balanced Nuclear Equation for the Alpha Emission of Bismuth-210:
[
\text{Bi-210} \rightarrow \text{He-4} + \text{Pb-206}
]
Explanation:
In nuclear decay processes, the nuclide undergoes transformations that involve either the emission of particles or the rearrangement of its components. In the case of alpha decay, an alpha particle (which is composed of 2 protons and 2 neutrons, symbolized as (\text{He-4})) is emitted from the nucleus of the decaying nuclide.
- Understanding the elements involved:
- Bismuth-210 ((\text{Bi-210})) has an atomic number of 83 and a mass number of 210. This means it contains 83 protons and (210 – 83 = 127) neutrons.
- The alpha particle ((\text{He-4})) has an atomic number of 2 and a mass number of 4, representing 2 protons and 2 neutrons.
- Effect of alpha decay on the nucleus:
- The emission of an alpha particle reduces the atomic number of the original nuclide by 2 and the mass number by 4. Therefore:
- The atomic number of Bismuth (Bi) decreases from 83 to (83 – 2 = 81).
- The mass number decreases from 210 to (210 – 4 = 206).
- The resulting product nuclide is Lead-206 ((\text{Pb-206})).
- Writing the nuclear equation:
- The alpha emission equation can be written as:
[
\text{Bi-210} \rightarrow \text{He-4} + \text{Pb-206}
] - This equation balances both the mass numbers and the atomic numbers:
- The mass number on both sides: (210 = 4 + 206).
- The atomic number on both sides: (83 = 2 + 81).
Thus, the correct nuclear equation for the alpha decay of bismuth-210 is (\text{Bi-210} \rightarrow \text{He-4} + \text{Pb-206}).