Question
A nucleus $$_n{X^m}$$ emits one $$\alpha $$ and two $$\beta $$-particles. The resulting nucleus is
A.
$$_n{X^{m - 4}}$$
B.
$$_{n - 2}{Y^{m - 4}}$$
C.
$$_{n - 4}{Z^{m - 4}}$$
D.
None of these
Answer :
$$_n{X^{m - 4}}$$
Solution :
The reaction can be shown as
Thus, the resulting nucleus is the isotope of parent nucleus and is $$_n{X^{m - 4}}.$$
The reaction can be shown as
Thus, the resulting nucleus is the isotope of parent nucleus and is $$_n{X^{m - 4}}.$$