I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Here’s what I got.
Strontium-90, ##””^90″Sr”##, undergoes beta minus decay, so your goal here will be to use what you know about the emission of a beta particle to figure out the resulting nuclide.
A beta particle, ##””_(-1)^(color(white)(aa)0)beta##, is simply a high-speed electron. When a radioactive nuclide undergoes beta minus decay, a neutron located inside its nucleus is being converted into a proton.
At the same time, a ##””_(-1)^(color(white)(aa)0)beta## particle and an electron antineutriono, ##bar(nu)_”e”##, are being emitted.
Now, because a neutron is being converted into a proton, the of the nuclide will not change. However, its , which tells you how many protons are present inside the nuclues, will change.
More specifically, the will increase by ##1##. A quick look in will show that strontium has an atomic number of ##38##.
This means that the daughter nuclide will have an atomic number of
##38 + 1 = 39##
Another quick look in the periodic table will reveal that the daughter nuclide is yttrium-90, ##””^90″Y”##.
You can thus write out the nuclear equation that describes the beta minus decay of strontium-90 like this
##color(green)(|bar(ul(color(white)(a/a)color(black)(“”_ 38^90″Sr” -> “”_ 39^90″Y” + “”_ (-1)^(color(white)(aa)0)beta + bar(nu)_”e”)color(white)(a/a)|)))##
Notice that mass and charge are conserved, since
##overbrace(90 = 90 + 0)^(color(blue)(“conservation of mass”))” “## and ##” “overbrace(38 = 39 + (-1))^(color(darkgreen)(“conservation of charge”))##
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