Question

Write balanced nuclear equations for (a) the loss of an alpha particle by Th-230. (b) the loss of a beta particle by lead-210. (c) the fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons. (d) the K-capture of Ar-37.

Short answer

Question: Write balanced nuclear equations for the following processes: (a) Alpha decay of Th-230 (b) Beta decay of Pb-210 (c) Fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons (d) K-capture of Ar-37 Answer: (a) ^{230}_{90}Th -> ^{4}_{2}He + ^{226}_{88}Ra (b) ^{210}_{82}Pb -> ^{0}_{-1}e + ^{210}_{83}Bi (c) ^{235}_{92}U -> ^{140}_{56}Ba + ^{93}_{36}Kr + 2^{1}_{0}n (d) ^{37}_{18}Ar + ^{0}_{-1}e -> ^{37}_{17}Cl

Step-by-step solution

(a) Alpha decay of Th-230

To write the balanced nuclear equation for the loss of an alpha particle by Th-230, we must make sure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal. Thorium-230 (Th-230) loses an alpha particle, which is a helium-4 nucleus, so we write: Th-230 -> He-4 + X Now we just need to find the nucleus X. An alpha particle consists of 2 protons and 2 neutrons, therefore, the mass number (A) decreases by 4, and the atomic number (Z) decreases by 2. The balanced nuclear equation for the alpha decay of Th-230 is: ^{230}_{90}Th -> ^{4}_{2}He + ^{226}_{88}Ra

(b) Beta decay of Pb-210

To write the balanced nuclear equation for the loss of a beta particle by lead-210 (Pb-210), we must ensure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal. Pb-210 loses a beta particle (electron). A beta particle has a mass number of 0 and an atomic number of -1: Pb-210 -> e-1 + X Now we just need to find the nucleus X. Since a beta particle has a mass number of 0, the mass number (A) remains unchanged. However, the atomic number (Z) increases by 1. The balanced nuclear equation for the beta decay of Pb-210 is: ^{210}_{82}Pb -> ^{0}_{-1}e + ^{210}_{83}Bi

(c) Fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons

To write the balanced nuclear equation for the fission of uranium-235 (U-235) to give barium-140 (Ba-140), another nucleus, and an excess of two neutrons, we must ensure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal: U-235 -> Ba-140 + Y + 2n Now we need to find the nucleus Y. Since the mass numbers (A) and atomic numbers (Z) must be equal on both sides of the equation, we can set up the following equations for mass number and atomic number: A: 235 = 140 + A_Y + 2(1) Z: 92 = 56 + Z_Y + 2(0) Solving for A_Y and Z_Y, we get: A_Y = 235 - 140 - 2 = 93 Z_Y = 92 - 56 = 36 So, the nucleus Y is ^{93}_{36}Kr. The balanced nuclear equation for the fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons is: ^{235}_{92}U -> ^{140}_{56}Ba + ^{93}_{36}Kr + 2^{1}_{0}n

(d) K-capture of Ar-37

To write the balanced nuclear equation for the K-capture of argon-37 (Ar-37), we must take into account that an electron from the inner shell is captured by the nucleus, and as a result, the atomic number (Z) decreases by 1, while the mass number (A) remains unchanged: Ar-37 + e -> X Now we just need to find the nucleus X. The mass number (A) remains the same; however, the atomic number (Z) decreases by 1. The balanced nuclear equation for the K-capture of Ar-37 is: ^{37}_{18}Ar + ^{0}_{-1}e -> ^{37}_{17}Cl

Key concepts

Alpha Decay

Alpha decay is a common type of radioactive decay where an unstable atomic nucleus emits an alpha particle. An alpha particle is similar to a helium nucleus and is composed of two protons and two neutrons.

In alpha decay, the original nucleus loses these four nucleons, resulting in a decrease in the mass number by four and a decrease in the atomic number by two. For example, when thorium-230 undergoes alpha decay, it emits an alpha particle and transforms into radium-226:

• Mass Number ( $A$): Decreases by 4.
• Atomic Number ( $Z$): Decreases by 2.

The equation for this transformation is: $${}_{90}^{230}\text{Th}{\to }_2^4\text{He}{+}_{88}^{226}\text{Ra}$$Alpha decay is essential in nuclear physics and contributes significantly to the understanding of radioactive elements.

Beta Decay

Beta decay is a process where a beta particle, which is an electron ( $e^{-}$), or a positron is emitted from an atomic nucleus. It involves the transformation of a neutron into a proton with the emission of an electron, or vice versa for beta plus decay.

In beta decay, the atomic number increases or decreases by one, while the mass number remains unchanged. For example, in the beta decay of lead-210:

• Mass Number ( $A$): Remains unchanged.
• Atomic Number ( $Z$): Increases by 1.

, Lead-210 transforms into bismuth-210 with the emission of an electron: $${}_{82}^{210}\text{Pb}{\to }_{-1}^0\text{e}{+}_{83}^{210}\text{Bi}$$This process helps change the element while keeping its mass the same and is a crucial concept in the study of radioactivity.

Nuclear Fission

Nuclear fission is a process where a large atomic nucleus splits into two or more smaller nuclei, along with the release of energy and often neutrons. It is a significant process used in both nuclear reactors and atomic bombs.

When uranium-235 undergoes fission, it splits into barium-140, krypton-93, and two neutrons in addition to releasing a huge amount of energy:

• Mass Number ( $A$): Sum of the products along with neutrons equals original mass number.
• Atomic Number ( $Z$): Sum of the atomic numbers of products equals original atomic number.

The balanced nuclear equation for this reaction is:$${}_{92}^{235}\text{U}{\to }_{56}^{140}\text{Ba}{+}_{36}^{93}\text{Kr}+2_0^1\text{n}$$Nuclear fission allows harnessing energy and is fundamental in understanding nuclear energy release.

K-Capture

K-capture, or K-electron capture, is a nuclear process where an electron from the closest electron shell is captured by the nucleus. This process reduces the atomic number by one but leaves the mass number unchanged.

During K-capture, the captured electron combines with a proton to form a neutron and a neutrino. For argon-37, the capture of a K-shell electron leads to its transformation into chlorine-37:

• Mass Number ( $A$): Remains unchanged.
• Atomic Number ( $Z$): Decreases by 1.

The balanced nuclear equation for argon-37 is:$${}_{18}^{37}\text{Ar}{+}_{-1}^0\text{e}{\to }_{17}^{37}\text{Cl}$$K-capture is critical in understanding how electron interactions affect nuclear properties and can lead to the stabilization of an atom.