A) \[6\alpha ,3\beta \]
B) \[3\alpha ,4\beta \]
C) \[4\alpha ,3\beta \]
D) \[3\alpha ,6\beta \]
Correct Answer: B
Solution :
Key Idea: \[\alpha -\]particle is the helium nucleus\[_{2}H{{e}^{4}}\] and\[\beta -\]particle is fast moving electron\[{{(}_{-1}}{{e}^{0}})\]When an\[\alpha -\]particle is emitted mass number decreases by 4 and atomic number by 2. Given, \[_{86}{{A}^{22}}{{\to }_{84}}{{B}^{210}}\] Decrease in mass number\[=222-210\] \[=12\] \[\therefore \]Number of \[\alpha -\]particles\[=\frac{12}{4}=3\] \[\therefore \]After 3\[\alpha -\]particle atomic number \[=86-3\times 2=80\] So, increase is atomic number from 80 to 84 is 4. Therefore,\[4\beta \] particles as 1 \[\beta -\]particle increases the atomic number by 1.
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