To determine which element Polonium-210 decays into after one alpha decay, we need to understand the process of alpha decay.
Alpha decay happens when an unstable nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which means it has an atomic number of 2 and a mass number of 4.
Given:
- The original atom: Polonium-210 ([tex]\(^{210}_{84}Po\)[/tex])
During alpha decay:
1. The polonium nucleus will emit an alpha particle ([tex]\(^{4}_{2}He\)[/tex]).
2. This emission will reduce the atomic number by 2 (since 2 protons are lost).
3. It will also reduce the mass number by 4.
So, let's calculate the new atomic number and mass number of the resulting atom after the decay:
1. New Atomic Number:
[tex]\[
84 \text{ (original atomic number of polonium)} - 2 = 82
\][/tex]
The new atomic number is 82.
2. New Mass Number:
[tex]\[
210 \text{ (original mass number of polonium)} - 4 = 206
\][/tex]
The new mass number is 206.
The element with an atomic number of 82 is lead (Pb). Therefore, the resulting atom after the alpha decay of Polonium-210 is Lead-206 ([tex]\(^{206}_{82}Pb\)[/tex]).
Given the options:
A. [tex]\(\quad{ }_{82}^{206} Pb\)[/tex]
B. [tex]\({ }^{210} Po\)[/tex]
C. [tex]\({ }_{86}^{214} Rn\)[/tex]
D. [tex]\({ }_{86}^{222} Rn\)[/tex]
E. [tex]\({ }_{82}^{214} Pb\)[/tex]
The correct answer is:
A. [tex]\(\quad{ }_{82}^{206} Pb\)[/tex]
