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Chapter 21: Problem 9

Why is it impossible for the isotope \({ }_{2}^{2}\) He to exist?

Short Answer

Expert verified
The isotope \({ }_{2}^{2}\) He does in fact exist. It is helium with 2 protons and no neutrons.

Step by step solution

01

Understanding isotopes

The isotope notation, \({ }_{Z}^{A}\) X, for an element X is defined as follows: the symbol X identifies the element, while the subscript Z and superscript A represent the atomic number and the atomic mass number of the atom, respectively. The atomic number Z defines the element itself (i.e., the number of protons), while the atomic mass number, A, is the sum of neutrons and protons in the nucleus.
02

Analyzing the given isotope

When you look at the isotope notation \({ }_{2}^{2}\) He, the superscript 2 refers to the atomic mass number A (sum of neutrons and protons), while the subscript 2, refers to the number of protons in helium (a defining property of helium). However, He is known to have two protons. The 2 protons define an atom as helium.
03

Conclusion about the impossibility

Considering that the atomic mass number A is 2 (sum of neutrons and protons), and it's known that helium has two protons, this would mean that there are no neutrons in this helium isotope. However, the element still exists as there are two protons which make it helium. Therefore, the isotope \({ }_{2}^{2}\) He does in fact exist, contradicting the premise of the question.

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Most popular questions from this chapter

A radioactive isotope of copper decays as follows: $$ { }^{64} \mathrm{Cu} \longrightarrow{ }^{64} \mathrm{Zn}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=12.8 \mathrm{~h} $$ Starting with \(84.0 \mathrm{~g}\) of \({ }^{64} \mathrm{Cu},\) calculate the quantity of \(^{64}\) Zn produced after 18.4 h.

How do nuclear reactions differ from ordinary chemical reactions?

Why do heavy elements such as uranium undergo fission while light elements such as hydrogen and lithium undergo fusion?

Consider this redox reaction: $$ \begin{array}{r} \mathrm{IO}_{4}^{-}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ \mathrm{I}_{2}(s)+\mathrm{IO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \end{array} $$ When \(\mathrm{KIO}_{4}\) is added to a solution containing iodide ions labeled with radioactive iodine- \(128,\) all the radioactivity appears in \(\mathrm{I}_{2}\) and none in the \(\mathrm{IO}_{3}^{-}\) ion. What can you deduce about the mechanism for the redox process?

Discuss the differences between a light water and a heavy water nuclear fission reactor. What are the advantages of a breeder reactor over a conventional nuclear fission reactor?
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