Question

In \(2006,\) an ex-KGB agent was murdered in London. The investigation following the agent's death revealed that he was poisoned with the radioactive isotope \({ }^{210} \mathrm{Po}\) which had apparently been added to his food. (a) \({ }^{210} \mathrm{Po}\) is prepared by bombarding \({ }^{209} \mathrm{Bi}\) with neutrons. Write an equation for the reaction. (b) The half-life of \({ }^{210} \mathrm{Po}\) is 138 days. It decays by \(\alpha\) particle emission. Write the equation for the decay process. (c) Calculate the energy of an emitted \(\alpha\) particle. Assume both the parent and daughter nuclei have zero kinetic energy. The atomic masses of \({ }^{210} \mathrm{Po},{ }^{206} \mathrm{~Pb},\) and \({ }_{2}^{4} \alpha\) are 209.98286 \(205.97444,\) and 4.00150 amu, respectively. (d) Ingestion of \(1 \mu \mathrm{g}\) of \({ }^{210}\) Po could prove fatal. What is the total energy released by this quantity of \({ }^{210}\) Po over the course of 138 days?

Short answer

(a) Bi-209 + n -> Po-210. (b) Po-210 -> Pb-206 + α. (c) 6.44 MeV. (d) 1.85 x 10^16 MeV.

Step-by-step solution

Determining the Reaction to Form Polonium-210

To form \({ }^{210} \mathrm{Po}\) through neutron bombardment, \({ }^{209} \mathrm{Bi}\) (bismuth) must capture a neutron (\(n\)) to form the isotope of interest. The nuclear reaction can be represented as follows: \(_{83}^{209}\mathrm{Bi} + _{0}^{1}n \rightarrow _{84}^{210}\mathrm{Po}\).This equation shows that \({ }^{209} \mathrm{Bi}\) gains one neutron, transforming into \({ }^{210} \mathrm{Po}\), increasing the atomic number by one to become polonium.

Writing the Alpha Decay Equation

In an alpha decay process, an \({ }^{210}\mathrm{Po}\) nucleus emits an alpha particle (\(\alpha\)), which consists of 2 protons and 2 neutrons, thereby transforming into a lead-206 (\({ }^{206}\mathrm{Pb}\)) nucleus. The equation for this decay process is:\[ { }^{210}_{84}\mathrm{Po} \rightarrow { }^{206}_{82}\mathrm{Pb} + { }^{4}_{2}\alpha \]This shows the parent polonium nucleus losing an alpha particle and decreasing its atomic number from 84 to 82, effectively transforming into lead.

Calculating Energy of the Emitted Alpha Particle

The energy of the emitted alpha particle can be calculated using the mass-energy equivalence principle (\(E=mc^2\)), where the mass defect (the difference in mass before and after decay) is converted into energy. The masses given are: - \({ }^{210}\mathrm{Po} = 209.98286\) amu - \({ }^{206}\mathrm{Pb} = 205.97444\) amu - \({ }^{4}_{2}\alpha = 4.00150\) amuCalculate the mass defect:\[\Delta m = (209.98286 - (205.97444 + 4.00150)) \text{ amu} = 0.00692 \text{ amu}\]Convert mass defect to energy using 1 amu = 931.5 MeV/c²:\[\Delta E = 0.00692 \text{ amu} \times 931.5 \frac{ \text{MeV}}{\text{amu}} = 6.44 \text{ MeV}\]Thus, the energy of the emitted alpha particle is approximately 6.44 MeV.

Calculating Total Energy Released by 1 µg of Polonium-210

First, determine the number of \(^{210}\mathrm{Po}\) atoms in 1 µg. The molar mass of \(^{210}\mathrm{Po}\) is roughly 210 g/mol. 1 gram contains:\[\frac{6.022 \times 10^{23} \text{ atoms/mol}}{210} = 2.87 \times 10^{21} \text{ atoms/g}\]For 1 µg (1 × 10⁻⁶ g), the number of atoms is:\[2.87 \times 10^{21} \times 10^{-6} = 2.87 \times 10^{15} \text{ atoms}\]Each decay releases 6.44 MeV, and thus, the total energy released is:\[2.87 \times 10^{15} \text{ atoms} \times 6.44 \text{ MeV/atom} = 1.85 \times 10^{16} \text{ MeV}\]Therefore, the total energy released over the course of 138 days by the ingestion of 1 µg of \(^{210}\mathrm{Po}\) is approximately \(1.85 \times 10^{16} \text{ MeV}\).

Key concepts

Radioactive Isotopes

Radioactive isotopes, also known as radioisotopes, are atoms with unstable nuclei. They decay over time, releasing energy in the form of radiation. This process is natural and occurs when the nucleus seeks a more stable configuration by emitting particles or electromagnetic waves.
One key feature of radioactive isotopes is their half-life. This is the time taken for half the quantity of the isotope to decay.

• Example: Polonium-210 ( ^{210} ext{Po} ), used in the described exercise, has a half-life of 138 days.
• The decay is predictable, allowing calculations of remaining quantities over time.

Understanding radioactive isotopes is crucial for safe handling and application in various fields, such as medicine or nuclear energy.

Alpha Decay

Alpha decay is a type of radioactive decay where an alpha particle is emitted from the nucleus of an atom. An alpha particle is composed of two protons and two neutrons (essentially a helium-4 nucleus).
When a nucleus undergoes alpha decay, the original atom's mass decreases by four units, and its atomic number drops by two.

• In the case of ^{210} ext{Po} , alpha decay transforms it into lead-206 ( ^{206} ext{Pb} ).
• The equation for this transformation is: ^{210}_{84} ext{Po} ightarrow ^{206}_{82} ext{Pb} + ^{4}_{2} ext{He}

This process alters the elemental identity of the atom, making it an interesting topic in nuclear chemistry.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus and can occur naturally or be induced artificially. These reactions result in the transformation of one element into another. One such artificial process is neutron bombardment, which is applicable in producing isotopes like ^{210} ext{Po}.
In the exercise, ^{209} ext{Bi} captures a neutron ( ^{1}_{0} ext{n} ) to become ^{210} ext{Po}:

• ^{209}_{83} ext{Bi} + ^{1}_{0} ext{n} ightarrow ^{210}_{84} ext{Po}
• This reaction signifies an increase in atomic number, converting bismuth into polonium.

Such reactions are foundational in nuclear chemistry, as they enable the creation of various isotopes for research and practical applications.

Mass-Energy Equivalence

Mass-energy equivalence is a fundamental principle in physics, articulated by Albert Einstein's famous equation E=mc^2 . It describes how mass can be converted into energy, and vice versa. This principle is essential for understanding nuclear energy release.
In the context of the exercise, the energy released during alpha decay is derived from the slight difference in mass (mass defect) before and after decay.

• The mass defect in the decay from ^{210} ext{Po} to ^{206} ext{Pb} was found to be 0.00692 amu.
• Using 1 ext{ amu} = 931.5 ext{ MeV} , the energy equivalence is calculated to be approximately 6.44 MeV.

This concept not only helps explain nuclear reactions but also provides insight into the vast potential energy stored within atomic nuclei.