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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