Problem 102 - Introductory Chemistry Atoms First

Question

Consider the radioactive decay of radium to radon: \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+?\) (a) Write the complete equation. (b) What type of decay is this? (c) Explain why radium-226 is likely to undergo the type of decay you named in part (b). (d) How much energy is released, in kilojoules, when 1 mole of \({ }^{226}\) Ra decays? [Molar masses: \({ }_{88}^{226} \mathrm{Ra}, 226.0254 \mathrm{~g} / \mathrm{mol} ;{ }_{86}^{222} \mathrm{Rn}, 222.0175 \mathrm{~g} / \mathrm{mol}\) \(\left.{ }_{2}^{4} \mathrm{He}, 4.0026036 \mathrm{~g} / \mathrm{mol}\right]\) (e) How much energy is released, in kilojoules, when \(1 \mathrm{~g}\) of \({ }_{88}^{226}\) Ra decays?

Step-by-Step Solution

Answer

The short answer to this question is as follows: (a) The complete equation for the decay is \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn} + { }_{2}^{4} \mathrm{He}\). (b) This decay process is an example of alpha decay. (c) Radium-226 is likely to undergo alpha decay due to the high number of protons in its nucleus, which creates significant repulsion forces that can be minimized by releasing an alpha particle. (d) The energy released when 1 mole of \({ }^{226}\mathrm{Ra}\) decays is approximately \(4.76 \times 10^9 \, \mathrm{kJ/mol}\). (e) The energy released when 1 gram of \({ }_{88}^{226}\) Ra decays is approximately \(2.11\times 10^7 \, \mathrm{kJ}\).

Step 1: (a) Write the complete equation

In order to write the complete equation for the radioactive decay of radium to radon, we must first identify the missing product. In this case, we see that radon has two fewer protons than radium, which indicates the emission of an alpha particle (helium nucleus). The complete equation is: \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn} + { }_{2}^{4} \mathrm{He}\)

Step 2: (b) What type of decay is this?

The decay process involves the emission of an alpha particle, a helium nucleus consisting of two protons and two neutrons. Therefore, this is an example of alpha decay.

Step 3: (c) Explain why radium-226 is likely to undergo this type of decay

Radium-226 is a heavy element with a high number of protons. In heavy elements, the repulsive forces between protons in the nucleus become progressively stronger as the numbers of protons increase. To minimize these forces, the nucleus may release an alpha particle, thereby decreasing the number of protons and stabilizing the nucleus. As a result, radium-226 is likely to undergo alpha decay.

Step 4: (d) How much energy is released, in kilojoules, when 1 mole of \({ }^{226}\) Ra decays?

To determine the energy released during the decay process, we first need to calculate the change in mass. The mass change can be determined by subtracting the mass of the decay products from the mass of radium-226: \(\Delta m = m_{\mathrm{Ra}} - (m_{\mathrm{Rn}} + m_{\mathrm{He}})\) Using the given molar masses: \(\Delta m = 226.0254 - (222.0175 + 4.0026036)\) \(\Delta m = 0.0052964 \, \mathrm{g/mol}\) Now we can calculate the total energy released using Einstein's mass-energy equivalence equation: \(E = \Delta m c^2\) where \(E\) is the energy released, \(c\) is the speed of light (\(2.9979\times10^8 \, \mathrm{m/s}\)), and \(\Delta m\) is the mass change in kilograms. We will need to convert the mass change from grams to kilograms: \(\Delta m = 0.0052964 \times 10^{-3} \, \mathrm{kg/mol}\) \(E = (0.0052964\times 10^{-3})(2.9979\times10^8)^2 \, \mathrm{J/mol}\) \(E = 4.76 \times 10^{12} \, \mathrm{J/mol}\) Finally, convert the energy to kilojoules: \(E = 4.76 \times 10^9 \, \mathrm{kJ/mol}\) Therefore, the energy released when 1 mole of \({ }^{226}\mathrm{Ra}\) decays is approximately \(4.76 \times 10^9 \, \mathrm{kJ/mol}\).

Step 5: (e) How much energy is released, in kilojoules, when \(1 \, \mathrm{g}\) of \({ }_{88}^{226}\) Ra decays?

To determine the energy released from the decay of 1 gram of radium-226, we first need to calculate the moles of radium in 1 gram: \(moles\, of\, radium = \frac{1\, \mathrm{g}}{226.0254 \, \mathrm{g/mol}}\) \(moles\, of \, radium = 0.00442\, \mathrm{mol}\) Now we can multiply the energy released per mole by the moles of radium in 1 gram: \(Energy = 0.00442 \, \mathrm{mol} \times 4.76 \times 10^9 \, \mathrm{kJ/mol}\) \(Energy = 2.11\times 10^7 \, \mathrm{kJ}\) Thus, the energy released when 1 gram of \({ }^{226}\mathrm{Ra}\) decays is approximately \(2.11\times 10^7 \, \mathrm{kJ}\).

Key Concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle to become more stable. An alpha particle consists of two protons and two neutrons, making it essentially a helium nucleus. This process is characteristic of heavy elements like radium, which undergo alpha decay to reduce their nuclear charge and mass. During alpha decay, the original element transforms into a new element with an atomic number that is reduced by two and a mass number decreased by four. This is why radium-226, when it undergoes alpha decay, turns into radon-222 along with the emission of an alpha particle.

• The released alpha particle is relatively heavy and low in penetration power.
• Despite being easy to block, it can be harmful if ingested or inhaled due to its ionizing capability.

Alpha decay is an important process because it can change an element into another, significantly impacting the element's chemical properties.

Radium-226

Radium-226 is a naturally occurring radioactive isotope of the alkaline earth metal radium, with an atomic number of 88. It is most famous for its emission of alpha particles as it decays into radon-222. Discovered by Marie and Pierre Curie, radium-226 is part of the uranium decay series and poses health risks due to its radioactivity. Radium-226 has an extensive half-life of about 1600 years, making it persist in the environment longer than many other radioactive substances.
This characteristic long half-life means that it poses long-term radiological hazards

• Radium-226 emits radiation that can cause damage to living tissues, leading to conditions such as cancer when exposure is substantial.
• Due to its radiative properties, it was historically used in luminescent paints for watch dials and instrument panels before its dangers were fully understood.

Understanding radium-226's decay helps us manage its applications and mitigate its risks effectively.

Mass-Energy Equivalence

Mass-energy equivalence is a principle articulated by Albert Einstein in his famous equation, \(E=mc^2\). According to this concept, mass can be converted into energy and vice versa, illustrating that the two are interchangeable. In the context of nuclear reactions, this principle helps in calculating the energy released during decay processes like that of radium-226.
The tiny mass loss during the decay of radium to radon translates into a significant amount of energy release, due to the multiplication factor of the speed of light squared (\(c^2\)) being enormous.

• The concept is foundational in nuclear physics as it provides a framework to understand how small mass changes can result in large energy outputs.
• In radioactive decay calculations, even slight discrepancies in mass can lead to substantial energy calculations.

Mass-energy equivalence is crucial for explaining the vast energy produced in atomic reactions, which has both beneficial applications like energy generation and detrimental effects like nuclear explosions.

Nuclear Stability

Nuclear stability is the ability of a nucleus to remain unchanged and not undergo radioactive decay. In atomic nuclei, the balance between the number of protons and neutrons determines this stability.
For heavy elements like radium-226, excessive protons lead to too much electrostatic repulsion between them, often resulting in instability. The nucleus attempts to attain a stable configuration through decay processes that reduce its mass and charge, such as alpha decay where radium-226 loses protons and neutrons.

• A stable nucleus optimizes attractive nuclear forces (through neutrons) to counter the repulsive electric forces between positively charged protons.
• Nuclei with an optimal neutron-to-proton ratio tend to be more stable.
• Radium-226, being heavy and naturally having a higher proton number, is prone to instability, explaining its tendency to decay into radon-222.

Nuclear stability is essential to understanding the behavior and transformation of radioactive substances in nature and how actively they change over time.