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jacob193

[tex]^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}[/tex]

Y is Po-218.

  • A = 218
  • Z = 84.

Explanation

[tex]^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}[/tex]

Here's the symbol of a particle in a nuclear reaction. [tex]^{A}_{Z}\text{Y}[/tex].

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

  • The mass number on the left hand side is 222.
  • The two mass numbers on the right hand side add up to A + 4.
  • 222 = A + 4.
  • A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

  • The atomic number on the left hand side is 86.
  • The two atomic number on the right hand side add up to __ + __.
  • 86 = __ + __.
  • Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as [tex]^{218}_{\phantom{2}84}\text{Po}[/tex]. Hence the equation:

[tex]^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}[/tex].

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