A certain element is radioactive. It emits alpha particles. Each of its nuclei contains 92 protons and 142 neutrons. Write a nuclear equation that describes the change that takes place by alpha emission?

1 Answer
Jun 21, 2016

Answer:

#""_ (color(white)(a)92)^234"U" -> ""_ (color(white)(a)90)^230"Th" + ""_2^4alpha#

Explanation:

The first thing to do here is make sure that you have a clear understanding of what an alpha particle is.

An alpha particles is simply the nucleus of a helium-4 atom, #""^4"He"#, and it contains #2# protons and #2# neutrons.

http://education.jlab.org/glossary/alphaparticle.html

Now, your next goal here is to figure out the mass number of the radioactive isotope given to you by the problem.

As you know, the mass number, #A#, is calculated by adding the number of protons, i.e. the atomic number, #Z#, and the number of neutrons, #N#, present in a nucleus.

#color(blue)(|bar(ul(color(white)(a/a)A = Z + Ncolor(white)(a/a)|)))#

In your case, you have #92# protons and #142# neutrons in the nucleus, which means that the nuclide's mass number will be

#A = "92 protons" + "142 neutrons" = 234#

Next, use isotope notation to write the isotopic symbol for this element

https://socratic.org/chemistry/nuclear-chemistry/isotope-notation

A quick look in the periodic table will reveal that your unknown element is uranium, #"U"#, and so you have #""_ (color(white)(a)92)^234"U"# as the isotope notation here.

Now, when a nuclide emits an alpha particle, two things happen

  • its atomic number decreases by #2# because an lpha particle contains #2# protons
  • its mass number decreases by #4# because an alpha particle contains #2# protons and #2# neutrons

You can thus set up your nuclear equation like this

#""_ (color(white)(a)92)^234"U" -> ""_m^n"X" + ""_2^4alpha#

In any nuclear reaction, charge and mass are conserved, which means that you can write

#92 = m + 2 " "->" "# conservation of charge

#234 = n + 4 " "->" "# conservation of mass

You will thus have

#m = 92 - 2 = 90" "# and #" "n = 234 - 4 = 230#

Another quick look in the periodic table will reveal that the element that has #90# protons in its nucleus is thorium, #"Th"#. The nuclide produced by the alpha decay will thus be thorium-230, #""_ (color(white)(a)90)^230"Th"#.

The balanced nuclear equation will thus be

#color(green)(|bar(ul(color(white)(a/a)color(black)(""_ (color(white)(a)92)^234"U" " "->" " ""_ (color(white)(a)90)^230"Th" " "+" " ""_2^4alpha)color(white)(a/a)|)))#