(a)
Interpretation:
Type of emission occurs in transition from 21082Pb to 21082Pb has to be determined.
Concept Introduction:
Alpha particle (α) represents nucleus of helium atom. It contains two protons and two neutrons. Emission of this alpha particle is called alpha decay. Symbol of alpha particle is 42He.
General equation for alpha decay is as follows:
AZX→A−4Z−2Y+42He
Here,
A is the mass or nucleon number.
Z is the atomic number.
X is the symbol of the element.
Beta emission (β−) process represents conversion of neutron into 1 proton and 1 electron. This proton stays within nucleus and electron is emitted out of atom. This electron is called the beta particle (0−1e).
The generic equation of the beta decay is as follows:
AZX→AZ+1Y+e0−1
Here,
A is the mass or nucleon number.
Z is the atomic number.
X and Y is the symbol of the element.
Gamma rays (γ) are energy photons with more energy than X-rays. Loss of gamma ray causes no change in mass number as well as atomic number.
(a)
Answer to Problem 10PE
Type of emission that occurs in transition from 21082Pb to 21082Pb is gamma emission.
Explanation of Solution
Skeleton equation for transition from 21082Pb to 21082Pb is as follows:
21082Pb→21082Pb+AZX
Since this transition cause no change in
21082Pb→21082Pb+energy
(b)
Interpretation:
Type of emissions occurs in transition from 23491Pa to 23089Ac and then 23089Ac to 23090Th have to be determined.
Concept Introduction:
Refer to part (a).
(b)
Answer to Problem 10PE
Type of emissions occurs in transition from 23491Pa to 23089Ac and then 23089Ac to 23090Th are alpha and beta emission respectively.
Explanation of Solution
Skeleton equation for transition from 23491Pa to 23089Ac is as follows:
23491Pa→23089Ac+AZX
Since transition causes loss in mass number by 4 and in atomic number by 2 hence 1 alpha particle is lost in this transition. Therefore alpha emission is taken place in this transition.
Hence resultant nuclear equation for alpha emission of 23491Pa is as follows:
23491Pa→23089Ac+42He
Skeleton equation for transition from 23089Ac to 23090Th is as follows:
23089Ac→23090Th+AZX
Since this transition causes atomic number to increase by 1 thus this transition is called beta emission. Resultant nuclear equation for beta emission of 23089Ac is as follows:
23089Ac→23090Th+0−1e
(c)
Interpretation:
Type of emissions occurs in transition from 23490Th to 23088Ra and then 23088Ra to 23088Ra have to be determined.
Concept Introduction:
Refer to part (a).
(c)
Answer to Problem 10PE
Type of emissions occurs in transition from 23490Th to 23088Ra and then 23088Ra to 23088Ra are alpha and gamma emission respectively.
Explanation of Solution
Skeleton equation for transition from 23490Th to 23088Ra is as follows:
23490Th→23088Ra+AZX
Since transition causes loss in mass number by 4 and in atomic number by 2 hence 1 alpha particle is lost in this transition. Therefore alpha emission is taken place in this transition.
Hence resultant nuclear equation for alpha emission of 23490Th is as follows:
23490Th→23088Ra+42He
Skeleton equation for transition from 23088Ra to 23088Ra is as follows:
23088Ra→23088Ra+AZX
Since this transition cause no change in atomic number as well as in mass number thus this transition is gramma emission. Resultant nuclear equation for gamma emission of 23088Ra is as follows:
23088Ra→23088Ra+energy
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Chapter 18 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
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