Question

if \(_{92} \mathrm{U}^{238}\) undergoes successively \(8 \alpha\) decays and $6 \beta$ decays then resulting nucleus is (A) \(\mathrm{Pb}^{206}\) (B) \(\mathrm{Pb}^{208}\) (C) \(\mathrm{Pb}^{214}\) (D) None of these

Short answer

The resulting nucleus after 8 alpha decays and 6 beta decays of \(_{92} \mathrm{U}^{238}\) is (A) \(\mathrm{Pb}^{206}\).

Step-by-step solution

Understanding Alpha decay and Beta decay

In an Alpha decay, an atom emits an alpha particle which consists of 2 protons and 2 neutrons. As a result, the atomic number (number of protons) of the decaying nucleus decreases by 2 and the mass number (number of protons + number of neutrons) decreases by 4. In a Beta decay (specifically, beta-minus decay), a neutron is converted into a proton, with the emission of an electron. The atomic number increases by 1, while the mass number remains the same.

Start with the given isotope

The initial isotope given is \(_{92} \mathrm{U}^{238}\). Initially, the atomic number (Z) is 92 and the mass number (A) is 238.

Calculate the result after 8 Alpha decays

For 8 alpha decays, there will be a loss of \(8 * 2 =16\) protons and \(8 * 4 =32\) nucleons (protons and neutrons combined). After these decays, the atomic number will be \(92 - 16 = 76\) and the mass number will be \(238 - 32 = 206\). Now, the nucleus is represented by \(_{76} \mathrm{X}^{206}\), where X is the symbol of the resulting element with atomic number 76.

Calculate the result after 6 Beta decays

For 6 beta decays, the mass number remains the same while the atomic number increases by 6. The updated atomic number will be \(76 + 6 = 82\). The nucleus after all decays will be \(_{82} \mathrm{Y}^{206}\), where Y is the symbol of the resulting element with atomic number 82.

Identify the resulting element

Based on the periodic table, the element with atomic number 82 is Lead (Pb). Thus, the resulting nucleus after 8 alpha and 6 beta decays will be \(_{82} \mathrm{Pb}^{206}\). The correct answer is: (A) \(\mathrm{Pb}^{206}\).