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anonymous

  • one year ago

In the following equation for the alpha decay of radium, which particle has been emitted? superscript 226 subscript 88 Ra yields superscript 222 subscript 86 Rn plus ? Visual: https://gyazo.com/35fcec654ccb09753641b2936a765075 1) Helium-4 2) Helium-2 3) A beta particle 4) A hydrogen nucleus and a beta particle

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  1. Rushwr
    • one year ago
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    What do u think ?

  2. Rushwr
    • one year ago
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    \[Ra ^{226}_{88} \rightarrow Ra ^{222}_{86} + He ^{4}_{2} +\nu \] As u can see there is a reduction of 4 units in atomic mass and reduction of 2 units in atomic number you can say that the emitted atom is He as it has an atomic mass of 4 and atomic number which is 2

  3. anonymous
    • one year ago
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    Is it Helium-4?

  4. Rushwr
    • one year ago
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    Did u get it ?

  5. Rushwr
    • one year ago
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    yep

  6. anonymous
    • one year ago
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    Great, thanks!

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