Question

A neutron-activated sample emits gamma rays at energies that are consistent with the decay of mercury-198 nuclei from an excited state to the ground state. If the reaction that takes place is$\mathrm{n}+(?) \rightarrow{ }^{198} \mathrm{Hg}^{*}+\mathrm{e}^{-}+\bar{v},$ what is the nuclide "(?)" that was present in the sample before neutron activation? ( \(\mathrm{W}\), tutorial: neutron detector)

Short answer

Answer: The unknown nuclide present in the sample before neutron activation was gold-197 (Au-197).

Step-by-step solution

Write down reaction equation

We are given the reaction equation as: $$\mathrm{n}+(?) \rightarrow{ }^{198}\mathrm{Hg}^{*}+\mathrm{e}^{-}+\bar{v}$$

Identify conservation laws

For reactions of this sort, the following conservation laws are typically applied: 1. Conservation of atomic number (Z) 2. Conservation of mass number (A)

Apply conservation of atomic number (Z)

The atomic number of the neutron is 0, and mercury's atomic number is 80. Let the atomic number of the unknown nuclide be \(Z\). Since the electron's atomic number is -1, the reaction equation in terms of atomic number conservation is: $$0 + Z = 80 - 1$$ Solving for \(Z\), we get: $$Z = 80 - 1$$ $$Z = 79$$

Apply conservation of mass number (A)

The mass number of the neutron is 1, and the mass number of the excited mercury-198 nucleus is 198. Let the mass number of the unknown nuclide be \(A\). The reaction equation in terms of mass number conservation is: $$1 + A = 198$$ Solving for \(A\), we get: $$A = 198 - 1$$ $$A = 197$$

Determine the unknown nuclide

Now that we have the atomic number (Z) and the mass number (A) for the unknown nuclide, we can determine its symbol. The element with atomic number 79 is gold (Au). Therefore, the unknown nuclide is Au-197. So the activation reaction is: $$\mathrm{n}+\mathrm{^{197}Au} \rightarrow{ }^{198}\mathrm{Hg}^{*}+\mathrm{e}^{-}+\bar{v}$$