Question

If \[{}_{13}^{27}Al\] is a stable isotope. \[{}_{13}^{29}Al\] is expected to disintegrate by
A. Alpha emission
B. beta emission
C. Positron emission
D. Proton emission

Answer 1

Hint: We have to remember that the conversion of one or two chemical elements or an isotope (or induced by bombarding it with an energetic particle) into another element (also known as nuclide) is called the nuclear reaction. This is also called nuclear transmutation. Aluminum atomic number is $13$ and aluminum has $22$ known isotopes, ranges from \[{}_{}^{22}Al\] to\[{}_{}^{43}Al\]. In $22$ known isotopes, \[{}_{13}^{27}Al\] is the only stable isotope.

Complete step by step answer:
We must remember that radioactive decay is a process, in which the unstable nucleus loses energy in the form of emitting radiation to form stable nuclides. The emitting radiation may be alpha particles, beta emission, positron emission, gamma rays and an electron capture. The original element and the newly formed nucleus after radiation is called parent and daughter nucleus respectively.
Beta ($\beta $) emission: When the nuclei contain too many neutrons, the nuclei undergo beta ($\beta $) emission. Neutrons are greater than protons. In beta emission the mass number is even the same after the decay process but the atomic number is increased to $ + 1$.
General reaction is,
\[{}_Z^AY \to {}_{Z + 1}^AX + {}_{{}_{ - 1}}^0\beta \]
\[{}_{13}^{29}Al\] is expected to disintegrate to give \[{}_{14}^{29}Si\] and the nuclear reaction is,
\[{}_{13}^{29}Al \to {}_{14}^{29}Si + {}_{{}_{ - 1}}^0\beta \]
Number of neutrons in aluminum is $16$ and after the emission number of neutrons is 15.
Option B. beta emission is the correct answer.
Now we discuss other options as,
Positron (${\beta ^ + }$) emission: In this decay the mass number remains the same and the atomic number decreases by one after the reaction. Therefore, the option C is incorrect.
Alpha ($\alpha $) emission: The atomic nucleus decays and emits alpha particles, with an atomic number is reduced by two and mass number is reduced by four. Therefore, the option A is incorrect.
Proton ($P$) emission: In this decay a proton is ejected from a nucleu

So, the correct answer is Option B.

Note: We have to remember that in nuclear reactions, protons and neutrons are not destroyed they are sifts around them. If the element contains the same atomic number (same number of protons) but different mass number (different number of neutrons) then that is called isotopes. For example: Hydrogen contains three isotopes, they are protium $\left( {{}_1^1H} \right)$, deuterium $\left( {{}_1^2H} \right)$, and tritium $\left( {{}_1^3H} \right)$.