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Chapter 18: Problem 2485

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

Short Answer

Expert verified
The final atomic number Z of the resulting nucleus after the given sequence of emissions is 74. Therefore, the correct answer is (C) 74.

Step by step solution

01

Initial Atomic Number

First, recognize that the initial atomic number Z is 92. This means that there are 92 protons in the nucleus.
02

Count Proton Change from Alpha Emissions

Count the number of alpha emissions: there are 10 alpha particles emitted. Each alpha particle causes the atomic number to decrease by 2, meaning a loss of 20 protons from the nucleus.
03

Count Proton Change from Beta-Minus Emissions

Count the number of beta-minus emissions: there are 5 beta-minus particles emitted. Each beta-minus particle causes the atomic number to increase by 1, meaning the addition of 5 protons to the nucleus.
04

Count Proton Change from Beta-Plus Emissions

Count the number of beta-plus emissions: there are 3 beta-plus particles emitted. Each beta-plus particle causes the atomic number to decrease by 1, meaning the loss of 3 protons from the nucleus.
05

Calculate Final Atomic Number

Combine the changes in the atomic number caused by all emissions: start with the initial atomic number (92 protons), subtract the protons lost in alpha emissions (20 protons) and beta-plus emissions (3 protons), and add the protons gained in beta-minus emissions (5 protons). Final Atomic Number = 92 - 20 + 5 - 3 = 74
06

Choose the Correct Option

The final atomic number Z of the resulting nucleus is 74, which corresponds to option (C).

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Most popular questions from this chapter

The binding energy Per nucleon of \({ }_{8} \mathrm{O}^{16}\) is $7.97 \mathrm{MeV}\( and that of \)_{8} \mathrm{O}^{17}\( is \)7.75 \mathrm{MeV}$ The energy \((\mathrm{in}-\mathrm{MeV})\) required to remove a neutron from $_{8} \mathrm{O}^{17}$ is (A) \(3.65\) (B) \(7.86\) (C) \(3.52\) (D) \(4.23\)

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