(c) 69 33 As decays by β^+ emission.

Name: Problem 2, Chapter 25: General Chemistry: Principles and Modern Applications
Uploaded: 2021-03-20T08:26:55-08:00
Duration: 6 min 35 s
Channel: Aadit Sharma
Description: (c) 69 33 As decays by β^+ emission.

Key Concepts

Nuclear Transmutation

Nuclear transmutation in the context of beta plus decay refers to the conversion of one chemical element into another as a result of the change in the number of protons within the nucleus. Although the mass number stays constant, the change in the proton number alters the identity of the element, playing a critical role in nuclear reactions and element formation.

Positron Emission

Positron emission is the specific mechanism involved in beta plus decay where a positron, which is the antimatter counterpart of the electron, is emitted. This emission helps proton-rich nuclei move towards a more energetically favorable, or stable, state by reducing the number of protons.

Beta Plus Decay

Beta plus decay is a type of radioactive decay in which a proton in the nucleus is converted into a neutron, emitting a positron and an electron neutrino. This process results in the reduction of the atomic number by one while the mass number remains unchanged, leading to the transformation of the parent nucleus into a daughter nucleus.

Transcript

00:01 Okay, this question has a total of four subparts.

00:07 So starting with part a, we have lead, which undergoes two successive beta emissions.

00:17 So lead, which is 214 pb and 82, goes to emission.

00:28 It goes from the first emission, we don't know what our product is.

00:34 But we also know that we have a beta emission so we also add beta particle because the ejected beta particle has a negative 1 charge which is equivalent to the atomic number of negative 1 the unknown product should have one extra proton so we know that lead p b which is 82 equals 82 plus 1 goes as 83.

01:11 The atomic number of 83 identifies as virulium b .i.

01:21 Since the mass number of unknown product remains the same because the beta emission or the beta particle has a zero mass, so the value of a will remain equals 214.

01:40 So the unknown product will be 214 beryllium 83 which undergoes beta emission.

01:53 So we have unknown product plus negative 1 beta.

02:00 And this is our second beta emission.

02:05 And the first beta emission respectively.

02:11 Again, similarly for the second beta emission, it will go as, so the atomic number will go as 83 plus 1, which goes as 84, and 84 atomic number goes to po.

02:31 The mass number will again remains the same.

02:35 So our final product obtained will be 214 po, which is polonium 84.

02:49 Using the same technique, we have to solve all the parts of this question...

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