Problem 58 - General Chemistry: Principles and Modern Applications

Question

The overall change in the radioactive decay of \({238}_{92} \mathrm{U}\)to 206 \(\mathrm{Pb}\) is the emission of eight \(\alpha\) particles. Show that if \(_{82}^{206} \mathrm{Pb}\)this loss of eight \(\alpha\) particles were not also accompanied by six \(\beta^{-}\) emissions, the product nucleus would still be radioactive.

Step-by-Step Solution

Answer

Without the six \(\beta^{-}\) emissions, eight \(\alpha\) decays from \(_{92}^{238} \mathrm{U}\) would result in \(_{76}^{206}\mathrm{Os}\), not \(_{82}^{206}\mathrm{Pb}\). Therefore, the product nucleus would still be radioactive, as osmium-206 would continue to decay until it becomes lead-206.

Step 1: Understand Alpha Emission

Alpha emission is a type of radioactive decay where an alpha particle is emitted from a nucleus. An alpha particle consists of 2 protons and 2 neutrons, essentially a helium-4 nucleus \(_{2}^{4}\mathrm{He}\). So, when a nucleus goes through an alpha decay process, it loses 2 protons and 2 neutrons - decreasing its atomic number by 2 and atomic mass by 4.

Step 2: Understand Beta Emission

Beta emission is another radioactive decay process. In beta-minus (\(\beta^{-}\)) decay specifically, a neutron in the nucleus changes into a proton and an electron. The electron is ejected as a beta particle, increasing the atomic number by 1 but leaving the atomic mass unchanged.

Step 3: Calculate the Change from Alpha Emissions only

Suppose uranium-238 only goes through alpha decays, emitting eight \(\alpha\) particles. Each alpha decay decreases the atomic number by 2 and the atomic mass by 4. So, after 8 alpha decays, the atomic number would decrease by \(8*2 = 16\) and the atomic mass would decrease by \(8*4 = 32\). The resulting isotope would be \(_{76}^{206}\mathrm{X}\). The atomic number 76 corresponds to the element osmium (Os), so the product isotope would be \(_{76}^{206}\mathrm{Os}\).

Step 4: Conclusion

Since the isotope produced is \(_{76}^{206}\mathrm{Os}\), and not \(_{82}^{206}\mathrm{Pb}\), which was the desired end point from the problem, your product is still radioactive and will continue decaying until it reaches \(_{82}^{206}\mathrm{Pb}\). Therefore, without the six \(\beta^{-}\) emissions, the product nucleus would still be radioactive.

Key Concepts

Alpha Emission

Alpha emission is a fascinating process in radioactive decay. It involves the discharge of an alpha particle from the nucleus of an atom. An alpha particle is actually a tightly bound clump consisting of two protons and two neutrons, essentially forming what we know as a helium-4 nucleus, represented as

• Alpha particles have a positive charge due to the presence of two protons.
• When an atom emits an alpha particle, its atomic number decreases by 2 and its mass number reduces by 4.

Let's use Uranium-238 ( ₉₂²³⁸U) as an example. When it undergoes alpha decay, its atomic number changes from 92 to 90, and the mass number shifts from 238 to 234. This results in the formation of Thorium-234 ( ₉₀²³⁴Th). This process continues, leading to significant transformation in the nucleus as a smaller, more stable element emerges.

Beta Emission

Beta emission is another type of nuclear decay, different from alpha emission. It typically involves the transformation of a neutron into a proton within the atom's nucleus. During this process, a beta particle, which is effectively an electron, is ejected. This changes the dynamics within the atom:

• The atomic number of the element increases by 1, due to the conversion of a neutron into a proton.
• However, the atomic mass remains the same because neutrons and protons have nearly the same mass.

In the example of Uranium-238 ( ₉₂²³⁸U), without ^beta emissions, the product of the alpha decay would continue to be radioactive. Through six beta-minus ( ^β⁻) emissions, such as in the decay of neutrally charged Thorium-234, an atomic transformation occurs. Each beta emission shifts the nucleus closer to a stable element, such as Lead-206 ( ₈₂²⁰⁶Pb), by increasing the atomic number without altering the mass.

Nuclear Decay Process

The nuclear decay process encompasses a series of transformations that an unstable nucleus undergoes to attain stability. This involves any combination of alpha and beta emissions, depending on the original isotope and its path to stability. A nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves:

• Alpha decay reduces both atomic mass and number, effectively decreasing the size and charge of the nucleus.
• Beta decay adjusts the ratio of protons to neutrons, altering the atomic number and promoting a stable nucleus.

Returning to Uranium-238 ( ₉₂²³⁸U), when it emits eight alpha particles, it gradually transforms into different elements, with each alpha or beta decay modifying its atomic structure. Yet, as the exercise illustrates, simply undergoing eight alpha emissions without any beta emissions will transform uranium into an unstable isotope, ₇₆²⁰⁶Osium, which would require additional decay. The beta emissions are critical; they adjust the neutron-to-proton ratio, eventually leading to the stable isotope of Lead-206 ( ₈₂²⁰⁶Pb) and consequently marking the end of the decay chain.