Radioactive decay

Answer:

Number of beta particles emitted = Initial number of atoms - Number of atoms remaining

Number of beta particles emitted = (8.0 x 10^12) - (2.0 x 10^12)

Number of beta particles emitted = 6.0 x 10^12

*Explanation: Each Cobalt-60 atom that decays emits one beta particle. Therefore, the number of beta particles emitted is equal to the number of Cobalt-60 atoms that have decayed.*

Answer:

* Spontaneous: The decay of a nucleus occurs on its own, without being triggered or influenced by any external factors such as temperature, pressure, or chemical reactions.
* Random: It is impossible to predict which specific nucleus in a sample will decay next, or when a particular nucleus will decay. The decay events occur randomly and follow statistical probabilities.

During alpha decay, the Po-210 nucleus emits an alpha particle (Helium nucleus). This results in the Po-210 nucleus changing to a Lead nucleus (Pb), a different element. The number of protons in the nucleus decreases by 2, and the number of neutrons decreases by 2, thus changing the element's identity.

Alpha decay involves the emission of an alpha particle, which consists of 2 protons and 2 neutrons, from the nucleus of an atom. The number of protons in the nucleus defines the element. Because alpha decay changes the number of protons (decreases by 2), the atom transforms into a different element with a different atomic number (number of protons).

Calculation:
Number of neutrons = Mass number - Number of protons = 238 - 92 = 146 neutrons.

Explanation:
Uranium-238 is radioactive because it has an excess of neutrons in its nucleus compared to the stable Uranium-235. This neutron excess makes the nucleus unstable, causing it to undergo radioactive decay.

1. The isotope may have an excess of neutrons in the nucleus.
2. The nucleus of the isotope may be too heavy (i.e., have a very high nucleon number).

Answer:

Alpha decay emits a helium nucleus (⁴He₂).

* Nucleon number (A): 241 - 4 = 237
* Proton number (Z): 95 - 2 = 93

Therefore, the daughter nucleus has a nucleon number of 237 and a proton number of 93.

Answer:

During beta-minus decay, a neutron within the nucleus transforms into a proton and an electron. The electron (β⁻ particle) is emitted from the nucleus. This results in an increase of the proton number by 1, while the nucleon number remains unchanged.

Beta decay often results in a more stable nucleus because it reduces the neutron-to-proton ratio. Nuclei with too many neutrons compared to protons are unstable. The decay converts a neutron into a proton, bringing the nucleus closer to the belt of stability on the nuclide chart.

Equation: $_{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He

Explanation: Alpha decay involves the emission of an alpha particle ($_{2}^{4}He$). The mass number decreases by 4 (238 to 234) and the atomic number decreases by 2 (92 to 90).

During beta-minus decay, a neutron in the nucleus transforms into a proton and an electron (beta-minus particle). The atomic number increases by 1, while the mass number remains the same.