A) \[\beta \]-rays
B) \[\alpha \]-rays
C) \[{{\beta }^{+}}\]-rays
D) \[\gamma \]-rays
Correct Answer: A
Solution :
\[\underset{less\,stable}{\mathop{_{11}N{{a}^{24}}}}\,\xrightarrow{{}}\underset{stable}{\mathop{_{11}N{{a}^{23}}}}\,+\underset{netutron}{\mathop{_{0}{{n}^{1}}}}\,\] The neutron on decomposition gives a prof and a \[\beta \]-particle \[{{(}_{-1}}{{e}^{0}})\] \[_{0}{{n}^{1}}\xrightarrow{{}}\underset{proton}{\mathop{_{1}{{H}^{1}}}}\,+\underset{\beta -particle}{\mathop{_{-1}{{e}^{0}}}}\,\] Hence, \[_{11}N{{a}^{24}}\] gets changed into \[_{11}N{{a}^{23}}\] means of \[\beta \]-emission.
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