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Chapter 18: Problem 2513

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

Expert verified
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

01

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.
02

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.
03

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.
04

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.
05

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) \(\mathrm{Pb}^{206}\).

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Most popular questions from this chapter

If the binding energy of electron in a hydrogen atom is \(13.6 \mathrm{eV}\), the energy required to remove the electron form the first state of \(\mathrm{Li}^{2+}\) is. (A) \(13.6 \mathrm{eV}\) (B) \(30.6 \mathrm{eV}\) (C) \(122.4 \mathrm{eV}\) (D) \(3.4 \mathrm{eV}\)

excited hydrogen atom emits a Photon of wave length \(\lambda\) in returning to the ground state The quantum number \(\mathrm{n}\) of excited state is given by (A) \(\sqrt{[}(\lambda . \mathrm{R}-1) /(\lambda \mathrm{R})]\) (B) \(\sqrt{[}(\lambda \mathrm{R}) /(\lambda \mathrm{R}-1)]\) (C) \(\sqrt{[\lambda R}(\lambda \mathrm{R}-1)]\) (D) \(\lambda \mathrm{R}(\mathrm{R}-1)\)

which of the following series in the spectrum of hydrogen atom lies in the visible legion of the electro magnetic spectrum? (A) Paschen (B) Lyman (C) Brakett (D) Balmer

The total energy of the electron in the first excited state of hydrogen is \(-3.4 \mathrm{eV}\). what is the kinetic energy of the electron in this state? (A) \(6.8 \mathrm{eV}\) (B) \(3.4 \mathrm{eV}\) (C) \(-3.4 \mathrm{eV}\) \((\mathrm{D})-6.8 \mathrm{eV}\)

The Rutherford revolution Per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of (A) \(10^{15}\) (B) \(10^{20}\) (C) \(10^{10}\) (D) \(10^{19}\)
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