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A nucleus $^m_nX$ emits one $\alpha$ particle and two $\beta^-$ particles. The resulting nucleus is
A
$^{m-4}_{n-2}Y$
B
$^{m-6}_{n-4}Z$
C
$^{m-6}_{n}Z$
D
$^{m-4}_{n}X$
Detailed Solution
An $\alpha$ particle is $^4_2He$: its emission decreases the mass number by 4 and the atomic number by 2.
After one $\alpha$ emission: mass number = m − 4, atomic number = n − 2.
A $\beta^-$ particle is an electron: its emission leaves the mass number unchanged and increases the atomic number by 1.
After two $\beta^-$ emissions: mass number = m − 4, atomic number = n − 2 + 2 = n.
The atomic number is again n, so the element is the same (an isotope of X).
The resulting nucleus is $^{m-4}_{n}X$.
After one $\alpha$ emission: mass number = m − 4, atomic number = n − 2.
A $\beta^-$ particle is an electron: its emission leaves the mass number unchanged and increases the atomic number by 1.
After two $\beta^-$ emissions: mass number = m − 4, atomic number = n − 2 + 2 = n.
The atomic number is again n, so the element is the same (an isotope of X).
The resulting nucleus is $^{m-4}_{n}X$.