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Consider the reaction below. What would the daughter nuclide be?

[tex]\[{}_{83}^{210} \text{Bi} \longrightarrow ? + {}_{-1}^{0} \text{e}\][/tex]

A. Hg-210
B. Po-210
C. Bi-210
D. Pb-210

Answer :

To solve the problem of finding the daughter nuclide in the nuclear reaction involving bismuth-210 ([tex]\(^{210}_{83}\text{Bi}\)[/tex]) where a beta particle ([tex]\(^{0}_{-1}e\)[/tex]) is emitted, follow these steps:

1. Understand the Reaction Type: This is a type of nuclear decay known as beta decay. In beta decay, a neutron in the nucleus is transformed into a proton and an electron (beta particle) is emitted.

2. Changes in Atomic Number: With the transformation of a neutron to a proton during beta decay, the atomic number of the original nuclide increases by 1. The mass number remains unchanged because the total number of nucleons (protons and neutrons combined) in the nucleus stays the same.

3. Apply the Changes:
- The original nuclide is bismuth-210 ([tex]\(^{210}_{83}\text{Bi}\)[/tex]).
- After beta decay, the atomic number [tex]\(83\)[/tex] increases by [tex]\(1\)[/tex], resulting in an atomic number of [tex]\(84\)[/tex].
- The mass number remains [tex]\(210\)[/tex].

4. Identify the Element: The element with an atomic number of [tex]\(84\)[/tex] is polonium (Po).

5. Write the Daughter Nuclide: Therefore, the resulting daughter nuclide is polonium-210 ([tex]\(^{210}_{84}\text{Po}\)[/tex]).

So, in conclusion, the daughter nuclide formed from the beta decay of bismuth-210 is polonium-210.

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