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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

Which of the following poses a greater health hazard: a radioactive isotope with a short half-life or a radioactive isotope with a long half-life? Explain. [Assume same type of radiation \((\alpha\) or \(\beta)\) and comparable energetics per particle emitted.

What makes water particularly suitable for use as moderator in a nuclear reactor?

These equations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify X. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (b) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{55}^{144} \mathrm{Cs}+{ }_{37}^{90} \mathrm{Rb}+2 \mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{87} \mathrm{Br}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{62}^{160} \mathrm{Sm}+{ }_{30}^{72} \mathrm{Zn}+4 \mathrm{X}\)

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.

Discuss factors that lead to nuclear decay.
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