A) 5 and 8
B) 6 and 8
C) 8 and 6
D) 8 and 10
Correct Answer: C
Solution :
Key Idea When a nucleus emits an \[\alpha \]-particle\[(_{2}^{4}He)\], the mass number decreases by 4 units and atomic number decreases by 2 units and by the loss of one \[\beta \]-particle atomic number increases by one unit but mass number does not change. The given nuclear charge is \[{{y}^{{{A}^{x}}}}\xrightarrow{{}}{{y}^{-{{10}^{x-32}}}}+{{m}_{2}}H{{e}^{4}}+{{n}_{-1}}{{e}^{0}}\] Here, the loss of mass is by 32 units. So, number of a-particles emitted \[=\frac{32}{8}=8\] By the emission of Sa-particles, the atomic number of the daughter element should become y -16. But it is y -10. Therefore, 6a-particle is also emitted in the nuclear charge. Hence, the value of m and n are 8 and 6 respectively.
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