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

1. Rushwr

What do u think ?

2. Rushwr

$Ra ^{226}_{88} \rightarrow Ra ^{222}_{86} + He ^{4}_{2} +\nu$ 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

Is it Helium-4?

4. Rushwr

Did u get it ?

5. Rushwr

yep

6. anonymous

Great, thanks!