Write balanced nuclear equations for the following: (a) Formation of \({ }_{22}^{48} \mathrm{Ti}\) through positron emission (b) Formation of silver- 107 through electron capture (c) Formation of polonium- 206 through \(\alpha\) decay

Positron emission is a form of radioactive decay where a proton within a nucleus converts into a neutron, and in the process, it releases a positron. A positron is the antimatter equivalent of an electron and has the same mass but a positive charge. When this occurs, the atomic number of the element decreases by one, but the mass number remains the same. For instance, in the formation of \(_{22}^{48} \mathrm{Ti}\) through positron emission, we start with an element, vanadium, which has one more proton (\(_{23}^{48} \mathrm{V}\)). The reaction can be represented as: \[ _{23}^{48} \mathrm{V} \rightarrow _{22}^{48} \mathrm{Ti} + (\beta^+) \] In this equation, a vanadium-48 nucleus emits a positron (\(\beta^+\)), converting a proton into a neutron and forming a titanium-48 nucleus. This type of decay is common in proton-rich nuclides and helps nuclear stability by reducing the number of protons.

Electron capture is another type of radioactive decay where an inner orbital electron is captured by the nucleus. This process converts a proton into a neutron and reduces the atomic number by one while maintaining the same mass number. For silver-107, the formation involves the capture of an electron (\(e^-\)) by cadmium-107: \[ _{48}^{107} \mathrm{Cd} + e^- \rightarrow _{47}^{107} \mathrm{Ag} \] Here, cadmium-107 captures an electron transforming one of its protons into a neutron, resulting in the formation of silver-107. Electron capture is commonly observed in neutron-deficient isotopes and helps increase nuclear binding by converting protons into neutrons. The process typically happens in inner electron shells, making the transformation more efficient and aiding in nuclear stability.

In alpha decay, a nucleus emits an alpha particle, which consists of two protons and two neutrons (effectively a helium nucleus, \( \mathrm{He}^4_2 \)). This emission decreases the atomic number by 2 and the mass number by 4. For the formation of polonium-206 through alpha decay: \[ _{86}^{210} \mathrm{Rn} \rightarrow _{84}^{206} \mathrm{Po} + _{2}^{4} \mathrm{He} \] In this case, radon-210 undergoes alpha decay, releasing an alpha particle and transforming into polonium-206. Alpha decay commonly occurs in heavy nuclei where the strong nuclear force is less effective in binding nucleons. The emission of an alpha particle reduces the positive charge and the total nucleon count, providing increased stability to the residual nucleus. Alpha particles are relatively heavy and carry substantial energy, making them highly ionizing but less penetrating compared to other forms of radioactive particles.