Problem 22 - Chemistry

Question

The radioactive isotope \(^{242} \mathrm{Cm}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \(^{242} \mathrm{Cm}\) through many transformations to end up as \(^{206} \mathrm{P} \mathrm{b}\) . In the complete decay series, how many \(\alpha\) and \(\beta\) particles are produced?

Step-by-Step Solution

Answer

In the complete decay series of \(^{242} \mathrm{Cm}\) to \(^{206} \mathrm{Pb}\), 9 alpha particles and 4 beta particles are produced.

Step 1: Identify the initial and final isotopes

In this exercise, we are given that the initial isotope is \(^{242} \mathrm{Cm}\), and the final isotope is \(^{206} \mathrm{Pb}\). The atomic number of the initial isotope (Cm) is 96, and the atomic number of the final isotope (Pb) is 82.

Step 2: Calculate the change in mass and atomic number

We need to find the change in mass and the change in atomic number during the complete decay process. Change in mass = Initial mass - Final mass = \(242 - 206 = 36\) Change in atomic number = Initial atomic number - Final atomic number = \(96 - 82 = 14\)

Step 3: Determine the number of alpha and beta particles produced

Let x be the number of alpha particles and y be the number of beta particles produced during the decay process. Since an alpha particle decreases the atomic mass by 4, the total decrease in atomic mass due to alpha particle production is \(4x\). As beta particles do not cause a change in mass, no change is due to beta particles in this case. Similarly, since an alpha particle decreases the atomic number by 2, the total decrease in atomic number due to alpha particle production is \(2x\). A beta particle increases the atomic number by 1, so the total increase in atomic number due to beta particle production is \(y\). Now we can write two equations: 1. Change in mass = \(4x\) 2. Change in atomic number = \(2x + y\) Using the values calculated in Step 2, we have: 1. 36 = \(4x\) 2. 14 = \(2x + y\)

Step 4: Solve for x and y

First, solve for x from the first equation: 36 = \(4x \Rightarrow x = \frac{36}{4} = 9\) Now substitute this value of x in the second equation: 14 = \(2(9) + y \Rightarrow 14 = 18 + y \Rightarrow y = -4\) Since we have a negative value for y, we made an error in understanding the decay process. Beta decay increases the atomic number (not decreases). Therefore, the correct second equation should be: 2. Change in atomic number = \(2x - y\) Now, use the new second equation and the value of x: 14 = \(2(9) - y \Rightarrow 14 = 18 - y \Rightarrow y = 4\)

Step 5: Present the final answer

We found that during the decay process of \(^{242} \mathrm{Cm}\) to \(^{206} \mathrm{Pb}\), the number of alpha particles produced is 9 and the number of beta particles produced is 4.

Key Concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle consists of two protons and two neutrons, which is the same as a helium-4 nucleus. This process results in a decrease in both the atomic mass and the atomic number of the original nucleus. When a nucleus undergoes alpha decay:

• The atomic mass decreases by 4 units since 4 nucleons are emitted.
• The atomic number decreases by 2, due to the loss of two protons.

In the exercise, alpha decay played a crucial role in the transformation of isotope Cm (Californium, with 96 protons) into Pb (Lead, with 82 protons), by releasing 9 alpha particles. Each alpha decay decreased the mass number, gradually bringing the mass down from 242 to 206.

Beta Decay

Beta decay is another form of radioactive decay that involves the emission of a beta particle. Unlike alpha particles, a beta particle is an electron or a positron that is emitted from the nucleus. This occurs when a neutron in the nucleus converts into a proton (beta-minus decay), emitting an electron and an antineutrino, or when a proton changes into a neutron (beta-plus decay), emitting a positron and a neutrino. In either case:

• During beta-minus decay, the atomic number increases by 1 as a neutron becomes a proton.
• The atomic mass, however, remains unchanged because the total number of nucleons stays the same.

In the exercise, 4 beta particles were emitted as a part of the decay series from Cm to Pb, adjusting the atomic number after several alpha emissions, without affecting the overall mass number.

Radioactive Isotopes

Radioactive isotopes, or radionuclides, are unstable isotopes that spontaneously release particles or energy, in a process called radioactive decay, to achieve a more stable state. Each element can have several isotopes, and these isotopes differ by the number of neutrons in their nuclei. Some key points about radioactive isotopes include:

• They decay at a constant rate, known as half-life, which is the time it takes for half of a sample to decay.
• The decay can result in the emission of alpha particles, beta particles, or gamma rays.
• Radioactive decay often leads to the formation of different elements or isotopes.

In the given problem, we observe the radioactive isotope Cm as it transforms through various stages of decay, ultimately stabilizing as Pb following numerous alpha and beta emissions.

Atomic Mass Change

The change in atomic mass occurs due to the emission of radioactive particles during decay processes. This concept is pivotal in understanding how isotopes form new elements through decay mechanisms such as alpha and beta decay. In alpha decay:

• The atomic mass changes significantly as each alpha particle emitted contains two protons and two neutrons, decreasing the mass of the parent isotope by 4 units.

Beta decay, by contrast:

• Does not affect the atomic mass number since it involves the conversion of a neutron into a proton or vice versa without a loss or gain of nucleons.

In the exercise, the atomic mass of Cm shifted from 242 to 206, which required specific observations and calculations to understand the series of alpha decays taking place, while beta decay adjusted the atomic number without altering mass, hence preserving the total nucleon count.