Question

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?

Answer 1

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?