Problem 43 - General Chemistry: Principles and Modern Applications

Question

One member each of the following pairs of radioisotopes decays by \(\beta^{-}\) emission, and the other by positron \(\left(\beta^{+}\right)\) emission: \((\mathrm{a})_{15}^{29} \mathrm{P}\) and \(_{15}^{33} \mathrm{P} ;(\mathrm{b}) \stackrel{120}{53} \mathrm{I}\) and \(_{53}^{134} \mathrm{I} .\) Which is which? Explain your reasoning.

Step-by-Step Solution

Answer

\((\mathrm{a})_{15}^{33} \mathrm{P}\) and \(_{53}^{134} \mathrm{I}\) decay by \(\beta^{-}\) emission, and \((\mathrm{a})_{15}^{29} \mathrm{P}\) and \(_{53}^{120} \mathrm{I}\) decay by \(\beta^{+}\) emission.

Step 1: Compare the \(\beta^{-}\) and \(\beta^{+}\) decay processes

In \(\beta^{-}\) decay, the atomic number increases by 1 while the mass number remains the same because a neutron in the nucleus is converted into a proton and an electron is emitted.In \(\beta^{+}\) decay, the atomic number decreases by 1, while the mass number remains the same because a proton in the nucleus is converted into a neutron and a positron (the \(\beta^{+}\)) is emitted.

Step 2: Apply the understanding to the first pair

Take the first pair \((\mathrm{a})_{15}^{29} \mathrm{P}\) and _{15}^{33} \mathrm{P}. Both have the same atomic number (15), but different mass numbers (29 and 33). Hence, both isotopes could be the result of \(\beta^-\) decay where the neutron to proton ratio is attempting to be restored by converting a neutron into a proton. Therefore \((\mathrm{a})_{15}^{33} \mathrm{P}\) decays by \(\beta^{-}\) emission and \((\mathrm{a})_{15}^{29} \mathrm{P}\) decays by \(\beta^{+}\) emission.

Step 3: Apply the understanding to the second pair

Take the second pair \(_{53}^{120} \mathrm{I}\) and \(_{53}^{134} \mathrm{I}\). Both have the same atomic number (53), but different mass numbers (120 and 134). Similarly, both isotopes could be undergoing \(\beta^-\) decay where the neutron to proton ratio is being restored by converting a neutron into a proton. Therefore \(_{53}^{134} \mathrm{I}\) decays by \(\beta^{-}\) emission and \(_{53}^{120} \mathrm{I}\) decays by \(\beta^{+}\) emission.

Key Concepts

Beta-Minus Decay

Beta-minus decay, commonly represented as \( \beta^{-} \) decay, is a type of radioactive decay where an unstable atom emits an electron and an antineutrino. This process occurs when there are too many neutrons in the nucleus, leading to an imbalance in the neutron-to-proton ratio. To correct this imbalance and move towards greater stability, a neutron is transformed into a proton, which increases the atomic number by one while the mass number remains unchanged.

During this transformation, the neutron essentially splits into three smaller particles: a proton, which stays in the nucleus, an electron, which is ejected from the atom as the beta particle, and an antineutrino, which carries away excess energy. As a result of \( \beta^{-} \) decay, the element changes into another element, located one position higher on the periodic table but with the same mass number.

Beta-Plus Decay

Beta-plus decay, or \( \beta^{+} \) decay, is another type of radioactive decay process where a positron, the antimatter counterpart of the electron, is emitted. \( \beta^{+} \) decay happens when a nucleus has too many protons, which can lead to an unstable ratio of protons to neutrons. To regain stability, a proton in the nucleus is converted into a neutron.

During \( \beta^{+} \) decay, the proton doesn't just disappear but transforms into a neutron, a positron, and a neutrino. The positron and neutrino are emitted from the nucleus, and the atomic number of the element decreases by one, moving the element one position lower on the periodic table. Interestingly, because a positron is involved instead of an electron, this type of decay is also known as positron emission.

Neutron to Proton Conversion

The conversion from a neutron to a proton is a central feature of \( \beta^{-} \) decay. The neutron, composed of one up quark and two down quarks, undergoes a transformation wherein one of its down quarks changes into an up quark. This subatomic change results in a proton, which has two up quarks and one down quark. It also leads to the creation of a beta particle (electron) and an antineutrino.

The neutron to proton conversion is crucial for maintaining a stable nucleus. Atoms with too many neutrons or an unstable ratio of neutrons to protons will undergo beta decay to achieve a more stable state. Understanding this conversion is key to grasping why certain elements undergo radioactive decay and how the elements transform as a result.

Atomic Number Variation in Decay

The atomic number variation in decay refers to the change in the number of protons within an atom's nucleus during radioactive decay processes. In \( \beta^{-} \) decay, the atomic number increases by one since a neutron is converted into a proton. Conversely, in \( \beta^{+} \) decay, the atomic number decreases by one due to the conversion of a proton into a neutron.

This variation is significant because it results in the formation of a different element. The atomic number is the defining characteristic of an element, so when it changes, the nature of the element changes as well. This is why isotopes that undergo \( \beta^{-} \) decay climb up one spot on the periodic table, while those that undergo \( \beta^{+} \) decay move down one spot. Understanding how atomic numbers change during decay processes is pivotal in studying nuclear chemistry and the transformation of elements.