Write balanced nuclear equations for the following transformations: (a) polonium-210 emits alpha particle; (b) neptunium-235 undergoes electron capture; (c) fluorine-18 emits beta particle; (d) carbon-14 decays by beta emission.

In an alpha decay, an unstable nucleus emits an alpha particle which consists of 2 protons and 2 neutrons, with a net charge of +2. The mass number reduces by 4 and the atomic number decreases by 2 as a result of the decay. Given, Polonium-210, which has an atomic number 84. Polonium-210 decaying and emitting an alpha particle can be represented as: \( ^{210}_{84}Po \to ^{4}_{2}He + ^{A}_{Z}X \) where A = Mass number of the product, Z = Atomic number of the product, X = symbol of the product nucleus. To balance the equation, we need to find the corresponding A and Z values. Balancing the mass numbers: 210 = 4 + A A = 206 Balancing the atomic numbers: 84 = 2 + Z Z = 82 The resulting nucleus, with atomic number 82, belongs to the element Lead (Pb) So, the balanced nuclear equation for Polonium-210 emitting an alpha particle is: \( ^{210}_{84}Po \to ^{4}_{2}He + ^{206}_{82}Pb \)

During electron capture, a nucleus with an excess of protons absorbs an inner shell electron and combines the protons with the electron to produce a neutron. The process reduces the atomic number by 1, while the mass number stays the same. Given, Neptunium-235, which has an atomic number 93. Neptunium-235 undergoing electron capture can be represented as: \( ^{235}_{93}Np + ^{0}_{-1}e \to ^{A}_{Z}X \) Balancing the mass numbers: 235 = A A = 235 Balancing the atomic numbers: 93 - 1 = Z Z = 92 The resulting nucleus, with atomic number 92, belongs to the element Uranium (U). So, the balanced nuclear equation for Neptunium-235 undergoing electron capture is: \( ^{235}_{93}Np + ^{0}_{-1}e \to ^{235}_{92}U \)

In beta decay (β- decay), a neutron is transformed into a proton and an electron (beta particle), the latter being emitted. This process increases the atomic number by 1, while the mass number stays the same. Given, Fluorine-18, which has an atomic number 9. Fluorine-18 emitting a beta particle can be represented as: \( ^{18}_{9}F \to ^{0}_{-1}e + ^{A}_{Z}X \) Balancing the mass numbers: 18 = A A = 18 Balancing the atomic numbers: 9 + 1 = Z Z = 10 The resulting nucleus, with atomic number 10, belongs to the element Neon (Ne). So, the balanced nuclear equation for Fluorine-18 emitting a beta particle is: \( ^{18}_{9}F \to ^{0}_{-1}e + ^{18}_{10}Ne \)

We have already discussed beta decay in the previous part (c). Here, we apply the same process for Carbon-14. Given, Carbon-14, which has an atomic number 6. Carbon-14 emitting a beta particle can be represented as: \( ^{14}_{6}C \to ^{0}_{-1}e + ^{A}_{Z}X \) Balancing the mass numbers: 14 = A A = 14 Balancing the atomic numbers: 6 + 1 = Z Z = 7 The resulting nucleus, with atomic number 7, belongs to the element Nitrogen (N). So, the balanced nuclear equation for Carbon-14 decays by beta emission is: \( ^{14}_{6}C \to ^{0}_{-1}e + ^{14}_{7}N \)