Fill in the blank in each of the following nuclear equations. You will need to enter either a form of radiation or an isotope. If it is the latter, you must fully identify the isotope in superscript/subscript notation?

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1 Answer
Jun 1, 2018

Answer:

A. #color(white)(l)_10^(23)"Ne" to color(white)(l)_11^(23)"Na" + e^-#

B. #color(white)(l)_94^(242)"Pu" to color(white)(l)_2^(4)"He" + color(white)(l)_92^(238)"U"#

C. #color(white)(l)_84^(204)"Po" + e^(-) to color(white)(l)_83^(204)"Pb"#

Explanation:

Equation A describes a neon-23 atom undergoing radioactive decay; a sodium atom of the same mass number #A=23# is formed as the product; however the proton number #Z# decrease by #1#. A drop in #Z# without any changes in #A# indicates that some weak interactions must have been taken place.

The beta-plus decay process, in which a neutron converts to a proton while emitting an electron, is likely to be accountable for this observation:

#color(white)(l)_0^(1)"n"^0 to color(white)(l)_1^(1)"p"^+ + e^(-)#

Optional: Check for the conservation of charge in this equation:

#"L.H.S."=0=1+(-1)="R.H.S."#

Note that this problem does not state the charge on each of the species. It is, therefore, necessary to check #A# and #Z# numbers when completing the equation.

#color(white)(l)_10^(23)"Ne" to color(white)(l)_11^(23)"Na" + ul(color(black)("x"))#

#"x"# should carry a nucleon number of #0# and a baryon number of #-1#; what is #x#?

In the second equation B., plutonium 242 undergoes alpha decay and emits an alpha particle (helium 4 nucleus #color(white)(l)_color(purple)(2)^color(navy)(4)"He"#.) As a result, the #"A"# number of the product nuclei shall be lower than the parent nuclei by #color(navy)(4)# and the #"Z"# number lower by #color(purple)(2)#.

#color(white)(l)_94^(242)"Pu" to color(white)(l)_2^(4)"He" + color(white)(l)_92^(238)"X"#

The product should, therefore, contain #238# nucleons and #92# protons. What is the symbol for this element?

In the third equation C., polonium 204 undergoes electron capture to convert a proton into a neutron.

#color(white)(l)_1^(1)"p"^(+) + e^(-) to color(white)(l)_0^(1)"n"^0#

... which is nearly the same as the beta decay process, reversed. As a weak interaction process, the nucleon number of the product will be identical to that of the parent nuclei; the proton number, however, will decrease by #1# as one of the protons got converted to a neutron.

#color(white)(l)_84^(204)"Po" + e^(-) to color(white)(l)_color(purple)(83)^(204)"X"#

which element does #"X"# represent?