226Ra88 decays by emitting three alpha particles and three beta particles. what's the resulting nuclei?
Is it 238Bi83?
(Number on left is mass # and on right is atomic #)
I don't think so.
88Ra226 ==>2He4 + 2He4 + 2He4 + -1e0 + -1e0 + -1e0 + X.
88-2-2-2+1+1+1 = 88-3=85At
(88 = 2+2+2-1-1-1 + X = 3+85)
226 -4-4-4 = 214
(226 = 4+4+4+X = 12 + 214
So I think the end product is 85At214. Check my thinking.
To determine the resulting nuclei after the decay of 226Ra88, we need to understand the process of alpha and beta decay.
Alpha decay involves the emission of an alpha particle, which consists of two protons and two neutrons, from the nucleus. This results in the atomic number decreasing by 2 and the mass number decreasing by 4. In beta decay, a neutron inside the nucleus is converted into a proton, electron (beta particle), and an antineutrino or a proton is converted into a neutron, positron (beta particle), and a neutrino. This results in the atomic number increasing by 1 while the mass number remains the same.
Let's break down the process step by step:
1. Start with 226Ra88, which means there are 226 nucleons (protons + neutrons) and the atomic number (number of protons) is 88.
2. The decay involves three alpha particles being emitted. Each alpha particle contains two protons and two neutrons, so a total of 6 protons and 6 neutrons are lost.
3. Subtraction: 226 - 6 = 220 (new mass number) and 88 - 6 = 82 (new atomic number).
4. Next, three beta particles are emitted. Each beta particle adds one proton (resulting in an increase by 3 protons) to the nucleus, resulting in the atomic number increasing by 3.
5. Addition: 82 + 3 = 85 (new atomic number); the mass number remains unchanged at 220.
Therefore, the resulting nuclei after the decay of 226Ra88, with the emission of three alpha particles and three beta particles, would be 220Bi85 and not 238Bi83 as you suggested.
It's important to note that 238Bi83 is not a stable nuclide and decays with a half-life of about 10^19 years to produce 234U92 through alpha decay.