Question

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

Answer 1

What do u think ?

Answer 2

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

Answer 3

Is it Helium-4?

Answer 4

Did u get it ?

Answer 5

yep

Answer 6

Great, thanks!