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Tritium contains one proton and two neutrons. There is no significant proton- proton repulsion present in the nucleus. Why, then, is tritium radioactive?

Short Answer

Expert verified
Tritium is radioactive due to its unstable 2:1 neutron-to-proton ratio, leading to beta decay.

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01

Understanding Tritium's Composition

Tritium is an isotope of hydrogen that contains one proton and two neutrons in its nucleus. This basic understanding is crucial as the neutron-to-proton ratio plays a key role in determining a nucleus's stability.
02

Neutron-to-Proton Ratio

The stability of a nucleus is often dictated by the neutron-to-proton ratio. In tritium, this ratio is 2:1. Stable light isotopes typically have a neutron-to-proton ratio very close to 1:1. Tritium has a higher ratio, which suggests potential instability.
03

Excess Neutron Influence

An excess of neutrons can destabilize a nucleus. In tritium, the two neutrons mean there is a higher likelihood of neutron decay processes, like beta decay, which can convert a neutron into a proton, electron, and neutrino to stabilize the nucleus.
04

Radioactivity through Decay

Tritium decays via beta decay due to its unstable neutron-to-proton ratio. In this process, one of the neutrons is transformed into a proton, releasing a beta particle (electron) and an antineutrino. This process is what makes tritium radioactive.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactivity
Radioactivity is the process by which unstable atomic nuclei release energy in the form of particles or electromagnetic waves. This spontaneous decomposition occurs because certain atomic nuclei are not stable. When a nucleus is unstable, it will seek to reach a more stable state.
This process can result in the emission of particles such as alpha particles, beta particles, or gamma rays. These emissions lead to the transformation of one element into another. For instance, in the case of tritium, radioactivity is manifested through beta decay.
Understanding radioactivity helps us learn about nuclear reactions inside atoms and how elements transform. It's important for applications ranging from medical treatments to power generation.
Isotopes
Isotopes are different forms of the same element, where each has the same number of protons but a different number of neutrons in their atomic nuclei. They are crucial in understanding elements and their behaviors.
Here's what you need to know about isotopes:
  • They have identical chemical properties but different nuclear properties.
  • Some isotopes are stable, while others are radioactive.
Tritium is a radioactive isotope of hydrogen. Whereas most hydrogen atoms have just one proton, tritium contains two neutrons along with its single proton.
The presence of extra neutrons can lead to instability, which explains why tritium exhibits radioactive properties even without substantial proton-proton repulsion.
Beta Decay
Beta decay is a type of radioactive decay where a beta particle (electron or positron) is emitted from an atomic nucleus. This occurs in nuclei with an excess of neutrons or protons, leading to a more balanced neutron-to-proton ratio.
During beta decay, one of the following transformations occurs:
  • A neutron turns into a proton, emitting an electron and an antineutrino.
  • A proton turns into a neutron, emitting a positron and a neutrino.
In the case of tritium, beta decay involves the transformation of a neutron into a proton. This nuclear rearrangement results in the emission of a beta particle and an antineutrino.
Beta decay is fundamental in nuclear physics, significantly affecting isotope stability. It transforms the atomic nucleus into a more stable state by adjusting the neutron and proton counts, making it a critical concept in understanding nuclear reactions.

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Most popular questions from this chapter

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}(89.907738 \mathrm{amu})\) : $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: \({ }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=64 \mathrm{~h}\) Zirconium- \(90(89.904703 \mathrm{amu})\) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times 10^{-4}\) amu.) (b) Starting with 1 mole of \({ }^{90} \mathrm{Sr}\), calculate the number of moles of \({ }^{90} \mathrm{Sr}\) that will decay in a year. (c) Calculate the amount of heat released (in kJ) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in part \((\mathrm{b}).\)

Given that: $$ \mathrm{H}(g)+\mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=-436.4 \mathrm{~kJ} / \mathrm{mol} $$ calculate the change in mass (in \(\mathrm{kg}\) ) per mole of \(\mathrm{H}_{2}\) formed.

Why is strontium-90 a particularly dangerous isotope for humans? The half-life of strontium-90 is 29.1 years. Calculate the radioactivity in millicuries of \(15.6 \mathrm{mg}\) of \({ }^{90} \mathrm{Sr}\)

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of \(\mathrm{A},\) and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

A wooden artifact has a \({ }^{14} \mathrm{C}\) activity of 18.9 disintegrations per minute, compared to 27.5 disintegrations per minute for live wood. Given that the half-life of \({ }^{14} \mathrm{C}\) is 5715 years, determine the age of the artifact.

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