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The ionization energy of hydrogen atom is \(13.6 \mathrm{eV}\). What will be the ionization energy of \(\mathrm{He}^{+}\)? (a) \(13.6 \mathrm{eV}\) (b) \(54.4 \mathrm{eV}\) (c) \(122.4 \mathrm{eV}\) (d) zero

Short Answer

Expert verified
The ionization energy of \(\mathrm{He}^+\) is \(54.4 \mathrm{eV}\) (option b).

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. For a hydrogen atom, this energy is given as \(13.6 \, \mathrm{eV}\). This is important because it helps us understand how changes to the nucleus or the electron affect the energy required.
02

Bohr's Model and Ionization Energy

We will use Bohr's model to solve this. According to Bohr's theory, the ionization energy is proportional to \(Z^2\), where \(Z\) is the atomic number. Hydrogen, with an atomic number \(Z = 1\), has an ionization energy of \(13.6 \, \mathrm{eV}\).
03

Calculating Ionization Energy for He鈦

The helium ion \(\mathrm{He}^+\) has an atomic number \(Z = 2\). Therefore, the ionization energy for \(\mathrm{He}^+\) is calculated using the ratio of their atomic numbers squared: \(Z^2_{\mathrm{He}^+} / Z^2_{\mathrm{H}} = 2^2 / 1^2 = 4\). Multiply this by the ionization energy of hydrogen: \(13.6 \, \mathrm{eV} \times 4 = 54.4 \, \mathrm{eV}\).

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Key Concepts

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

Bohr's Model
Bohr's Model provides us with a simple and clear way to understand the atom. It was introduced by Niels Bohr in 1913 and focuses on the hydrogen atom. The main idea is that electrons orbit the nucleus in distinct paths called "orbits" or "shells." Each orbit has a fixed energy level. These levels are quantized, meaning electrons can only exist at specific energy levels. They cannot exist in between these levels.
This model also introduces the concept of ionization energy鈥攖he energy needed to remove an electron completely from an atom. It's crucial for calculating how much energy is needed to change an atom from its ground state to a more excited state or to remove an electron entirely. In Bohr's model, the ionization energy is dependent upon the atomic number squared, which allows for calculations extremely vital in chemistry, such as for our helium ion example.
In summary, Bohr's model, with its electron orbits and levels, greatly simplifies the atomic structure imagery and has paved the way for modern atomic physics.
Atomic Number
The atomic number is fundamentally important to understanding an element's identity. It represents the number of protons found in the nucleus of an atom and defines the element itself. For example, any atom with an atomic number of 1 is hydrogen, while one with an atomic number of 2 is helium. The more protons, the higher the atomic number and the heavier the element.
Furthermore, the atomic number determines an atom's position on the periodic table and influences its properties. In chemistry, it鈥檚 often used to calculate related properties, such as ionization energy. This is crucial for the helium ion \( \text{He}^{+} \), which with an atomic number of 2, plays a pivotal role in calculating its ionization energy using Bohr鈥檚 model. The ionization energy changes with atomic number, illustrating how tightly the electrons are bound to the nucleus.
It is important because it affects how atoms interact with each other, forming the basis of chemical reactions and bonding.
Helium Ion (He鈦)
The helium ion \( \text{He}^{+} \) is a positively charged ion of helium. This occurs when a helium atom loses one of its two electrons. The loss of an electron results in a net positive charge. In essence, the helium ion has an atomic number of 2, indicating two protons in its nucleus.
Helium ions like \( \text{He}^{+} \) are significant in understanding atomic interactions and ionization principles. Their atomic structure makes them a helpful example when learning about ionization. When calculating the ionization energy of \( \text{He}^{+} \), the process involves understanding the impact of the higher atomic number compared to hydrogen. Employing Bohr's model, we see that since \( \text{He}^{+} \) has an atomic number of 2, its ionization energy is four times that of hydrogen, resulting in 54.4 eV.
Thus, helium ions provide a strong basis for learning how ionization energies can be influenced by factors such as atomic number, making it an essential concept in physical chemistry.

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Most popular questions from this chapter

Which of the following statements is /are correct? (a) The energy of an electron is largely determined by its principal quantum number. (b) The energy of electron in an orbital in the hydrogen atom depends upon the value of principal quantum number only. (c) The value of principal quantum number for \(24^{\text {* }}\) electron is 3 . (d) The principal quantum number is a measure of the most probable distance of finding the electron around the nucleus

The number of electrons, neutrons and protons in a species are equal to 10,8 and 8 respectively. The proper symbol of the species is (a) \({ }^{16} \mathrm{O}_{\mathrm{g}}\) (b) \({ }^{18} \mathrm{O}_{8}\) (c) \({ }^{18} \mathrm{Ne}_{10}\) (d) \({ }^{16} \mathrm{O}_{8}^{-2}\)

Rutherford's experiment, which established the nuclear model of the atom, used a beam of (a) \(\beta\)-particles, which impinged on a metal foil and got absorbed (b) \(\gamma\)-rays, which impinged on a metal foil and ejected electrons (c) Helium atoms, which impinged on a metal foil and got scattered (d) Helium nuclei, which impinged on a metal foil and got scatterd.

Which of the following factors induce larger angle of deflection in Rutherford's scattering experiment? (a) The initial kinetic energy of striking \(\alpha\)-particles must be large (b) The initial kinetic energy of striking \(\alpha\)-particles must be small (c) The nucleus to which the \(\alpha\)-particles are striking must have small atomic number (d) The nucleus to which the \(\alpha\)-particles are striking must have high atomic number

Among the following series of transition metal ions, the one where all metal ion have \(3 \mathrm{~d}^{2}\) electronic configuration is (a) \(\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}\) (b) \(\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}\) (c) \(\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\) (d) \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\) (At. wt \(\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25)\)

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