Problem 14 - Chemistry and Global Awareness

Question

Write a balance reaction for the decay Ir-192. This isotope undergoes alpha and gamma emission at the same time.

Step-by-Step Solution

Answer

The balanced reaction is \( ^{192}_{77}\text{Ir} \rightarrow ^{188}_{75}\text{Re} + ^{4}_{2}\text{He} + \gamma \).

Step 1: Identify the Parent Isotope

The given parent isotope is Iridium-192, denoted as \(^{192}_{77}\text{Ir}\). This tells us that the isotope has a mass number (A) of 192 and an atomic number (Z) of 77.

Step 2: Determine Alpha Emission

Alpha emission involves the release of an alpha particle, \(^{4}_{2}\text{He}\), from the nucleus. During this process, the mass number decreases by 4 and the atomic number decreases by 2. Therefore, Iridium-192 becomes \(^{188}_{75}\text{Re}\) (Rhenium).

Step 3: Account for Gamma Emission

Gamma emission does not affect the atomic number or the mass number because it involves the release of high-energy photons, which carry off energy but no charge or mass. The isotope's identity as \(^{188}_{75}\text{Re}\) remains unchanged by gamma emission.

Step 4: Write the Balanced Nuclear Equation

Combining the changes, the balanced reaction for the decay of Iridium-192 can be expressed as: \[ ^{192}_{77}\text{Ir} \rightarrow ^{188}_{75}\text{Re} + ^{4}_{2}\text{He} + \gamma \] This equation accounts for the alpha and gamma emissions, maintaining balance in terms of atomic and mass numbers.

Key Concepts

Alpha Emission

Alpha emission is a type of nuclear decay where an atom's nucleus releases an alpha particle. Alpha particles consist of 2 protons and 2 neutrons, which forms a helium-4 nucleus symbolized as \(^{4}_{2}\text{He}\). During alpha emission, there is a significant change in the parent isotope. It loses these 4 nucleons, causing the mass number to decrease by 4 and the atomic number to decrease by 2.

For instance, when Iridium-192 undergoes alpha emission, it releases an alpha particle. As a result, it changes into an isotope of Rhenium (Re), lowering its mass number from 192 to 188 and its atomic number from 77 to 75. This transformation demonstrates how alpha emission alters both the identity and structure of an isotope.

Gamma Emission

Gamma emission is another form of nuclear decay, but it differs from alpha emission because it does not involve the loss of particles from the nucleus. Instead, gamma emission is the release of gamma rays, which are high-energy photons. These photons carry off excess energy without changing the atomic or mass numbers of the nucleus.

In the case of Iridium-192 undergoing decay, the gamma emission does not alter it further once it transforms into Rhenium through alpha emission. Gamma rays help dissipate excess energy from the nucleus, but they do not affect its composition. As such, the isotope remains as Rhenium-188, unchanged in terms of protons, neutrons, or electrons after gamma emission occurs.

Balanced Nuclear Equation

A balanced nuclear equation represents the transformation occurring in a nuclear reaction, maintaining the conservation of atomic and mass numbers. Writing a balanced equation for nuclear decay involves accounting for particles like alpha and gamma emissions.

To balance an equation, the sum of mass numbers and atomic numbers on both sides must be equal. For the decay of Iridium-192, this means starting from \(^{192}_{77}\text{Ir}\), with the emission of an alpha particle \(^{4}_{2}\text{He}\), transforming into \(^{188}_{75}\text{Re}\) and accounting for gamma emission (denoted as \(\gamma\)).

The balanced nuclear equation is:
\[^{192}_{77}\text{Ir} \rightarrow ^{188}_{75}\text{Re} + ^{4}_{2}\text{He} + \gamma\]
This balanced equation ensures that the mass number sums to 192 and the atomic number sums to 77, before and after the reaction, thereby satisfying the conservation laws of nuclear reactions.