Question

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

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

Initial Atomic Number

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

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.

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.

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.

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

Choose the Correct Option

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