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Chapter 18: Problem 2491

A radio-active nucleus \({ }^{\mathrm{A}} \mathrm{Z} \mathrm{X}\) emits $3 \alpha$ -particles and 2 Positrons. the ratio of number of neutron to that of Protons in the final nucleus will be (A) \([(\mathrm{A}-\mathrm{Z}-8) /(\mathrm{Z}-8)]\) (B) \([(\mathrm{A}-\mathrm{Z}-12) /(\mathrm{Z}-4)]\) (C) \([(\mathrm{A}-\mathrm{Z}-4) /(\mathrm{Z}-8)]\) (D) \([(A-Z-4) /(Z-2)]\)

Short Answer

Expert verified
The short answer for the problem is: The ratio of the number of neutrons to protons in the final nucleus after emitting 3 alpha particles and 2 positrons is given by (C) \(\frac{(A - Z) - 4}{Z - 8}\).

Step by step solution

01

Determine the initial composition

In the beginning, we have a nucleus represented by \({ }^{\mathrm{A}} \mathrm{Z} \mathrm{X}\). Here, A is the mass number, and Z is the atomic number. The mass number is the sum of the number of protons and neutrons, and the atomic number is the number of protons. We can use these definitions to express the initial number of protons and neutrons. Number of initial protons (P_initial) = Z Number of initial Neutrons (N_initial) = A - Z
02

Analyze the emissions

In the given decay process, the nucleus emits 3 alpha particles and 2 positrons. Let's see how these emissions affect the composition of the nucleus. 1. Emission of an alpha particle: An alpha particle consists of 2 protons and 2 neutrons. Therefore, when an alpha particle is emitted, the nucleus loses 2 protons and 2 neutrons. 3 alpha particles mean there is a loss of 6 protons and 6 neutrons. 2. Emission of a positron: A positron is an anti-particle of an electron, and when it is emitted, a proton changes into a neutron. Therefore, when a positron is emitted, the nucleus loses 1 proton and gains 1 neutron. 2 positrons mean there is a loss of 2 protons and a gain of 2 neutrons.
03

Calculate the final composition

Now, let's find out the final number of protons and neutrons in the nucleus after all emissions. Number of final protons (P_final) = P_initial - 6 (from alpha particles) - 2 (from positrons) = Z - 8 Number of final neutrons (N_final) = N_initial - 6 (from alpha particles) + 2 (from positrons) = (A - Z) - 4
04

Find the ratio of neutrons to protons

The ratio of the number of neutrons to protons in the final nucleus can now be determined: Ratio = N_final / P_final = \(\frac{(A - Z) - 4}{Z - 8}\) The correct answer is (C).

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Most popular questions from this chapter

The half life time of a radioactive elements of \(\mathrm{x}\) is the same as the mean life of another radioactive element \(\mathrm{y}\). Initially they have same number of atoms, then (A) \(\mathrm{y}\) will decay faster then \(\mathrm{x}\) (B) \(\mathrm{x}\) will decay faster then \(\mathrm{y}\) (C) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate at all time (D) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate initially.

The nucleus at rest disintegrate into two nuclear parts which have their velocities in the ratio \(2: 1\) The ratio of their nuclear sizes will be (A) \(2^{(1 / 3)}: 1\) (B) \(1: 2^{(1 / 3)}\) (C) \(3^{(1 / 2)}: 1\) (D) \(1: 3^{(1 / 2)}\)

The hydrogen atom can give spectral lines in the series Lyman, Balmer and Paschen. which of the following statement is correct (A) Lyman series is in the infrared region (B) Balmer series is in the ultraviolet region (C) Balmer series is in the visible region (D) Paschen series is in the visible region.

In the following disintegration series ${ }_{92} \mathrm{U}^{238} \rightarrow^{\alpha} \mathrm{x} \rightarrow^{\beta-}{ } \mathrm{Z} \mathrm{y}^{\mathrm{A}}$ The value of \(Z\) and \(A\) respectively will be (A) 90,234 (B) 92,236 (C) 88,234 (D) 91,234

A nucleus with \(\mathrm{Z}=92\) emits the following sequence $\alpha, \alpha, \beta^{-}\( \)\beta^{-}, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha\( The \)\mathrm{Z}$ of the resulting nucleus is (A) 76 (B) 78 (C) 74 (D) 82
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