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

Short Answer

Expert verified
\({ }_{2}^{2}\)He cannot exist due to the absence of neutrons to provide the necessary binding energy to hold the protons together against their electromagnetic repulsion.

Step by step solution

01

Understanding Atomic Structure

Helium (He) is an element in the periodic table with atomic number 2. This means in its nucleus, it has 2 protons. Any atom of helium must contain 2 protons in its nucleus in order to be helium. The superscript number, 2 in this case, indicates the mass number of the isotope, which is the sum of protons and neutrons in the nucleus.
02

Analysing the number of neutrons

If the atomic mass of the isotope is 2, and helium has 2 protons, then the remaining mass units must be the number of neutrons. In this case, \({ }_{2}^{2}\)He would have 0 neutrons.
03

Checking the stability

Generally, isotopes may be stable or unstable based on the proton-to-neutron ratio. However, having 0 neutrons would not provide the necessary nuclear binding energy to hold the protons together in the nucleus. Protons, having similar positive charges, repel each other due to the electromagnetic force. It's the strong force, provided by neutrons (which neither repel or attract protons and each other electromagnetically), that keeps the nucleus stable. In the absence of neutrons, as in this case, the nucleus would fall apart due to the electrostatic repulsion between the protons.
04

Conclusion

Hence, the isotope \({ }_{2}^{2}\)He cannot exist because its nucleus contains 2 protons and no neutrons to provide the necessary binding energy to hold the protons together against their electromagnetic repulsion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Atomic Structure
Atomic structure is fundamental to understanding the nature of elements and their isotopes. An atom is composed of a nucleus and an outer shell of electrons. The nucleus contains positively charged protons and neutrally charged neutrons, while the electrons are negatively charged and orbit around the nucleus. The number of protons, termed the atomic number, defines the element. For instance, helium has an atomic number of 2, which means any atom with two protons is recognized as helium.

The mass number of an isotope is given by adding the number of protons and neutrons within the nucleus. Helium's mass number in the discussed isotope is two, corresponding to its two protons. However, the absence of neutrons is unusual and deviates from the general structure of stable atomic nuclei, which typically include one or more neutrons.
Neutron Function in Nucleus
Neutrons play a pivotal role in the stability of an atom's nucleus. While they do not contribute to the chemical properties of an element, which are determined by electrons, neutrons provide several critical nuclear functions. They increase the mass of an atom without adding to its charge, and significantly, they mitigate the electrostatic repulsion between protons.

In addition to contributing to the overall mass, neutrons facilitate the strong nuclear force, which is the key to holding the nucleus together. Without neutrons, as is the case with the hypothetical isotope \( {}_{2}^{2} \)He, the protons would repel each other, leading to an unstable or non-existent nucleus.
Strong Nuclear Force
The strong nuclear force is one of the four fundamental forces of nature and is vital for the stability of atomic nuclei. This force acts between the nucleons (protons and neutrons) in the nucleus, providing the necessary glue that counteracts the repulsive electromagnetic force between like-charged protons.

It is the strongest of the four fundamental forces, but it operates over very short ranges, roughly limited to the distances between neighboring nucleons. The strong nuclear force is approximately 100 times stronger than the electromagnetic force, ensuring that nuclei stay intact despite the repulsive forces trying to tear them apart. Neutrons are essential contributors to this force, which helps to explain why an all-proton nucleus, such as \( {}_{2}^{2} \)He, cannot exist.
Proton-to-Neutron Ratio
The proton-to-neutron ratio is an indicator of isotopic stability. For low atomic number elements, stable isotopes will usually have a proton-to-neutron ratio near unity, meaning the number of protons and neutrons are relatively balanced. As elements get heavier, a higher neutron proportion is generally required to maintain a stable nucleus, due to the increasing electrostatic repulsion between the additional protons.

The ratio also affects the nuclear binding energy, which is the energy needed to hold the nucleus together. For example, the non-existence of the isotope \( {}_{2}^{2} \)He indicates a problem with this ratio, as it has two protons and no neutrons, which is insufficient to overcome proton repulsion and maintain stability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The radioactive potassium- 40 isotope decays to argon- 40 with a half-life of \(1.2 \times 10^{9}\) yr. (a) Write a balanced equation for the reaction. (b) A sample of moon rock is found to contain 18 percent potassium-40 and 82 percent argon by mass. Calculate the age of the rock in years.

Two radioactive isotopes \(\mathrm{X}\) and \(\mathrm{Y}\) have the same molar amount at \(t=0 .\) A week later, there are four times as many \(\mathrm{X}\) as there are \(\mathrm{Y} .\) If the half-life of \(\mathrm{X}\) is \(2.0 \mathrm{~d}\), calculate the half-life of \(\mathrm{Y}\) in days.

As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine-131 in a person's body is \(7.4 \mathrm{mCi}\) \(\left(1 \mathrm{mCi}=1 \times 10^{-3} \mathrm{Ci}\right) .\) Use the relationship rate \(=\lambda N\) to calculate the number of atoms of iodine- 131 to which this radioactivity corresponds. (The halflife of \({ }^{131} \mathrm{I}\) is 8.1 d. \()\).

The radius of a uranium-235 nucleus is about \(7.0 \times 10^{-3} \mathrm{pm} .\) Calculate the density of the nucleus in \(\mathrm{g} / \mathrm{cm}^{3}\). (Assume the atomic mass is 235 amu.)

(a) Assuming nuclei are spherical in shape, show that its radius \((r)\) is proportional to the cube root of mass number \((A) .(\mathrm{b})\) In general, the radius of a nucleus is given by \(r=r_{0} A^{\frac{1}{3}},\) where \(r_{0},\) the proportionality constant, is given by \(1.2 \times 10^{-15} \mathrm{~m}\). Calculate the volume of the \({ }^{238} \mathrm{U}\) nucleus.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free