A) mass number reduces by 2
B) mass number reduces by 6
C) atomic number reduces by 2
D) atomic number remains unchanged
Correct Answer: D
Solution :
Key Idea: In every type of emission whether it is \[\alpha \]-decay or \[\beta \]-decay, the mass number and atomic number of resultant nucleus depend on the decay which is being done. The \[\alpha \]-particle can be represented as \[_{2}H{{e}^{4}}\] and \[\beta \]-particle as \[_{-1}{{\beta }^{0}}\]. So, after emission of one \[\alpha \]-particle the mass number of resultant nucleus decreases by 4 unit and atomic number by 2 unit. Similarly after emission of one \[\beta \]-particle the atomic number increase by 1 unit keeping its mass number same. So, according to reaction (assuming \[_{z}{{X}^{A}}\] the initial nucleus) \[_{Z}{{X}^{A}}{{\to }_{Z-2}}{{Y}^{A-4}}{{+}_{2}}H{{e}^{4}}\] (\[\alpha \]-particle) and \[_{Z-2}{{Y}^{A-4}}{{\to }_{Z}}{{X}^{A-4}}+2{{(}_{-1}}{{\beta }^{0}})\] (\[2\beta \]-particles) So, by one \[\alpha \] and two \[\beta \]-emissions the atomic number remains unchanged. While mass number decreases by 4.
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