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How to position given radioactive element on the band of stability

Submitted by Anonymous (not verified) on Thu, 01/13/2011 - 16:34

1) Considering the position of Ra-218 in the band of stability, ONLY AN ALPHA PARTICLE will be emitted. 

Is this statement true or false?  Please explain why.  If false, change the capitalized words to make it true.

2) Write the balanced nuclear equations for the radioactive decay of U-238 to U-234.

(I wrote 238/92 U-------->234/90U +4/2He.  My teacher wrote"Th?"  Where am I wrong and what does Thorium have to do with this?

218/88Ra would become 214/86Rn if it lost an alpha particle. 218/88Ra has a  n/p ratio of 130/88 = 1.48 while 214/86Rn has a  n/p ratio of 128/86 = 1.49. This increase in n/p ratio is not helping stability.
However some atoms can convert a neutron into a proton and emit a beta particle - an electron (there are other ways - I'm not an expert here)
So 218/88Ra could lose a neutron and gain a proton, so mass number stays the same but atomic number increases by 1 and the atom becomes 218/89Ac which has a n/p ratio of
129/89 = 1.45 making it more stable. I think that it should be changed to only a beta particle will be emitted

When an atom loses an alpha particle it loses 4 units of mass and 2 units of atomic number because an alpha particle is 2 neutrons and 2 protons. So its mass = 4 and its atomic number = 2
So when 238/92 U loses an alpha particle, its mass number becomes 234 and its atomic number becomes 90. As only uranium atoms have atomic number 92, so the new atom is now another element with atomic number 90. So this is an atom of Thorium which has atomic number 90.
238/92 U-------->234/90Th +4/2He.