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Chapter 22: Problem 19

(a) Give the names and chemical symbols for the three isotopes of hydrogen. (b) List the isotopes in order of decreasing natural abundance. (c) Which hydrogen isotope is radioactive? (d) Write the nuclear equation for the radioactive decay of this isotope.

Short Answer

Expert verified
The three isotopes of hydrogen are: 1. Protium (\(^1H\)): most abundant with 99.98% natural abundance, 2. Deuterium (\(^2H\)): 0.015% natural abundance, and 3. Tritium (\(^3H\)): least abundant and radioactive. The nuclear equation for Tritium's radioactive decay is: \(^3H \rightarrow ^3He + e^- + \bar{\nu}_e\).

Step by step solution

01

Provide the names and chemical symbols of hydrogen isotopes

There are three isotopes of hydrogen, which are: 1. Protium (\( ^1H \)): It has one proton and no neutrons. 2. Deuterium (\( ^2H \)): It has one proton and one neutron. 3. Tritium (\( ^3H \)): It has one proton and two neutrons.
02

List the isotopes in order of decreasing natural abundance

The isotopes of hydrogen can be listed in order of decreasing natural abundance as follows: 1. Protium (\( ^1H \)): It is the most abundant isotope, with around 99.98% of natural abundance. 2. Deuterium (\( ^2H \)): It is less abundant than Protium, with around 0.015% of natural abundance. 3. Tritium (\( ^3H \)): It is the least abundant isotope, as it is only found in trace amounts. Its natural abundance is extremely low, around 10^-18 %.
03

Identify the radioactive isotope of hydrogen

Tritium (\( ^3H \)) is the only radioactive isotope of hydrogen.
04

Write the nuclear equation for the radioactive decay of the radioactive isotope

Tritium (\( ^3H \)) undergoes beta decay, in which it transforms one of its neutrons into a proton and emits an electron (beta particle) and an antineutrino. The nuclear equation for the radioactive decay of Tritium can be written as follows: \( ^3H \rightarrow ^3He + e^- + \bar{\nu}_e \) Here, \( ^3H \) is the tritium nucleus, \( ^3He \) is the helium-3 nucleus, \( e^- \) is the emitted electron (beta particle), and \( \bar{\nu}_e \) is the emitted antineutrino.

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Most popular questions from this chapter

Write a balanced equation for each of the following reactions: (a) Burning magnesium metal in a carbon dioxide atmosphere reduces the \(\mathrm{CO}_{2}\) to carbon. (b) In photosynthesis, solar energy is used to produce glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and \(\mathrm{O}_{2}\) from carbon dioxide and water. (c) When carbonate salts dissolve in water, they produce basic solutions.

Complete and balance the following equations: (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (c) \(\mathrm{MnO}_{2}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{AlP}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{Na}_{2} \mathrm{~S}(s)+\mathrm{HCl}(a q) \longrightarrow\)

SOLUTION Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88 . The complete chemical symbol for radium- 226 is therefore \({ }_{85}^{226} \mathrm{Ra}\). An alpha particle is a helium-4 nucleus, and so its symbol is \({ }_{2}^{4} \mathrm{He}\). The alpha particle is a product of the nuclear reaction, and so the equation is of the form $$ { }_{8}^{226} \mathrm{Ra} \longrightarrow{ }_{2}^{A} \mathrm{X}+{ }_{2}^{4} \mathrm{He} $$ where \(A\) is the mass number of the product nucleus and \(Z\) is its atomic number. Mass numbers and atomic numbers must balance, so $$ 226=A+4 $$ and $$ 88=Z+2 $$ Hence, $$ A=222 \text { and } Z=86 $$ Again, from the periodic table, the element with \(Z=86\) is radon (Rn). The product, therefore, is \({ }_{86}^{222} \mathrm{Rn}\), and the nuclear equation is $$ { }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} $$

The \(\mathrm{SF}_{5}^{-}\)ion is formed when \(\mathrm{SF}_{4}(g)\) reacts with fluoride salts containing large cations, such as \(\mathrm{CsF}(s)\). Draw the Lewis structures for \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{5}^{-}\), and predict the molecular structure of each.

Predict whether each of the following oxides is acidic, basic, amphoteric, or neutral: (a) \(\mathrm{NO}_{2}\), (b) \(\mathrm{CO}_{2}\), (c) \(\mathrm{Al}_{2} \mathrm{O}_{3}\), (d) \(\mathrm{CaO}\).
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