Radium-226 does not decay by beta decay. It decays by alpha decay to radon-222.
Radium 226 decays by alpha emission to Radon 222. A helium nucleus is emitted by alpha emission which makes the mass reduce by 4 and its atomic number by 2.
Thorium-230 is radioactive because it undergoes alpha decay, turning into radium-226 with the release of an alpha particle. This decay process is characteristic of radioactive elements.
226 Ra 88 ---> 225 Ac 89 +W boson W boson ---> e- + neutron
The equation for alpha decay from radium-226 (Ra-226) can be represented as follows: [ \text{Ra-226} \rightarrow \text{Rn-222} + \alpha ] In this equation, radium-226 (Ra-226) emits an alpha particle (α), which is essentially a helium nucleus, resulting in the formation of radon-222 (Rn-222). This process decreases the atomic number by two and the mass number by four.
Radium-226 does not decay by beta decay. It decays by alpha decay to radon-222.
Radium 226 decays by alpha emission to Radon 222. A helium nucleus is emitted by alpha emission which makes the mass reduce by 4 and its atomic number by 2.
The equation for the alpha decay of 226Ra: 88226Ra --> 86222Rn + 24He The alpha particle is represented as a helium (He) nucleus.
Thorium-230 is radioactive because it undergoes alpha decay, turning into radium-226 with the release of an alpha particle. This decay process is characteristic of radioactive elements.
By alpha decay Th-230 is transformed in Ra-226.
226 Ra 88 ---> 225 Ac 89 +W boson W boson ---> e- + neutron
Radium-226--------------------Radon-222 + alpha
When radium-226 undergoes alpha decay, it becomes radon-222. We write the equation like this: 88226Ra => 24He + 86222Rn Here we see the alpha particle written as a helium-4 nucleus, which is, in point of fact, what it is. Notice that the numbers that are subscripted are equal on both sides of the equation, and the superscripted numbers are as well. They must balance for your equation to be correct.
The equation for alpha decay from radium-226 (Ra-226) can be represented as follows: [ \text{Ra-226} \rightarrow \text{Rn-222} + \alpha ] In this equation, radium-226 (Ra-226) emits an alpha particle (α), which is essentially a helium nucleus, resulting in the formation of radon-222 (Rn-222). This process decreases the atomic number by two and the mass number by four.
Thorium-230 decays into radium-226 when it emits an alpha particle.
In alpha decay, the nucleus emits an alpha particle (helium nucleus) consisting of 2 protons and 2 neutrons. Thallium-230 undergoes alpha decay to produce an alpha particle (helium-4 nucleus) and become lead-226. The balanced nuclear equation for this process is: ([^{230}{81}Tl \rightarrow ^{4}{2}He + ^{226}_{82}Pb]).
The decay is: Th-230-------------Ra-226 + He-4