How is the atomic mass changed by the alpha decay?
Answer
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Hint:Alpha decay (also denoted by \[\alpha - \] decay) is a type of radioactive decay. In this an atomic nucleus emits an \[\alpha - \] particle (i.e., helium nucleus) and thereby transforms into a different atomic nucleus. The mass number that is reduced by four and an atomic number that is reduced by two. It generally occurs, when a massive nuclei has too large proton to neutron ratio.
Complete answer:
\[\alpha - \] decay involves the emission of an \[\alpha - \] particle by a radioactive isotope of an element. An alpha particle is helium whose atomic number is two and has a molar mass of four in the nucleus, \[{}_2^4He\] , which contains two protons and two neutrons. When the alpha particle is emitted, the atomic number of the atom decreases by two, since the atomic number is the number of protons, which causes the transmutation of the original element into a different element. The original atom's mass number decreases by four since the mass number is the sum of protons and neutrons in the nucleus.
In an \[\alpha - \] particle, it is having the stable configuration of particles with its two protons and two neutrons. \[\alpha - \] radiation reduces the ratio of protons to neutrons in the parent nucleus which brings the more stable configuration.
Considering an example of alpha decay, when Polonium- \[210\] having the atomic number \[84\] get decayed by the emission of alpha particles. This \[{}_{84}^{210}Po\,\]has \[84\] protons and \[126\] neutrons. The ratio of protons to neutrons is \[Z/N\; = {\text{ }}84/126\] , or \[Z/N\; = {\text{ }}0.667\]. A\[{}_{82}^{206}Pb\] nucleus has \[82\] protons and \[124\] neutrons which results in the ratio of \[Z/N\; = {\text{ }}82/124\] , or \[Z/N\; = {\text{ }}0.661\] . This small change in the \[Z/N\] ratio is enough to put the nucleus into a more stable state. This brings the "daughter" nucleus (decay product) into the region of stable nuclei in the Chart of the Nuclides.
\[{}_{84}^{210}Po\, \to \,{}_{82}^{206}Pb\, + \,{}_2^4He\]
Note:In alpha decay, the atomic number changes, so the parent (original) atoms and the daughter (decayed) atoms are different elements and therefore have different chemical properties. In this uranium- \[238\] gives thorium \[{}_{90}^{234}Th\] and helium \[{}_2^4He\] .
\[{}_{92}^{238}U\, \to \,{}_{90}^{234}Th\, + \,{}_2^4He\]
Complete answer:
\[\alpha - \] decay involves the emission of an \[\alpha - \] particle by a radioactive isotope of an element. An alpha particle is helium whose atomic number is two and has a molar mass of four in the nucleus, \[{}_2^4He\] , which contains two protons and two neutrons. When the alpha particle is emitted, the atomic number of the atom decreases by two, since the atomic number is the number of protons, which causes the transmutation of the original element into a different element. The original atom's mass number decreases by four since the mass number is the sum of protons and neutrons in the nucleus.
In an \[\alpha - \] particle, it is having the stable configuration of particles with its two protons and two neutrons. \[\alpha - \] radiation reduces the ratio of protons to neutrons in the parent nucleus which brings the more stable configuration.
Considering an example of alpha decay, when Polonium- \[210\] having the atomic number \[84\] get decayed by the emission of alpha particles. This \[{}_{84}^{210}Po\,\]has \[84\] protons and \[126\] neutrons. The ratio of protons to neutrons is \[Z/N\; = {\text{ }}84/126\] , or \[Z/N\; = {\text{ }}0.667\]. A\[{}_{82}^{206}Pb\] nucleus has \[82\] protons and \[124\] neutrons which results in the ratio of \[Z/N\; = {\text{ }}82/124\] , or \[Z/N\; = {\text{ }}0.661\] . This small change in the \[Z/N\] ratio is enough to put the nucleus into a more stable state. This brings the "daughter" nucleus (decay product) into the region of stable nuclei in the Chart of the Nuclides.
\[{}_{84}^{210}Po\, \to \,{}_{82}^{206}Pb\, + \,{}_2^4He\]
Note:In alpha decay, the atomic number changes, so the parent (original) atoms and the daughter (decayed) atoms are different elements and therefore have different chemical properties. In this uranium- \[238\] gives thorium \[{}_{90}^{234}Th\] and helium \[{}_2^4He\] .
\[{}_{92}^{238}U\, \to \,{}_{90}^{234}Th\, + \,{}_2^4He\]
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