A) \[6\alpha ,3\]
B) \[3\alpha ,4\]
C) \[4\alpha ,3\]
D) \[3\alpha ,6\]
Correct Answer: B
Solution :
\[{}_{86}{{A}^{222}}\to {}_{84}{{B}^{210}}\] Decrease in mass lumber = 222 - 210 = 12 \[\therefore \]Number of \[\alpha \] particles emitted \[=\frac{12}{4}=3\] \[{}_{86}{{A}^{222}}\to {}_{80}{{A}^{210}}\to {}_{84}{{B}^{210}}\] Increase in atomic number = 84 - 80 = 4 \[\therefore \] Number of \[\beta \]-particles emitted = 4
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