The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 neutrons. Which atomic symbol could represent an isotope that undergoes radioactive decay to produce Pb-206

Respuesta :

Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

  • Alpha particle emission  \left \{ {{y=4} \atop {x=2}}He \right.
  • Beta particle emission    \left \{ {{y=0} \atop {x=-1}}e \right.
  • gamma radiation             \left \{ {{y=0} \atop {x=0}}γ \right.

in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right +  \left \{ {{y=0} \atop {x=0}}γ\right.

Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right

Answer:

C

Explanation:

i think its c but im not for sure, sorry:(