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The radius of the second orbit of \(\mathrm{H}\) -atom is equal to the radius of (a) second orbit of \(\mathrm{He}^{+}\) ion (b) third orbit of \(\mathrm{Li}^{2+}\) ion (c) fourth orbit of \(\mathrm{He}^{+}\) ion (d) fourth orbit of \(\mathrm{Be}^{3+}\) ion

Short Answer

Expert verified
The radius of the second orbit of an H-atom is equal to the radius of the fourth orbit of a \(Be^{3+}\) ion.

Step by step solution

01

Understanding the Bohr Model

In the Bohr model of the hydrogen atom, the radius of the electron's orbit increases with the square of the principal quantum number (n). The radius of the nth orbit is given by the formula: \(r_n = n^2 \frac{{h^2}}{{4 \pi^2 m_e k e^2}}Z^{-1}\), where \(h\) is the Planck's constant, \(m_e\) is the electron mass, \(k\) is the Coulomb's constant, \(e\) is the electric charge of an electron, and \(Z\) is the atomic number.
02

Comparing the Radius of the Second Orbit of H-atom

We are given that the radius of the second orbit of the Hydrogen atom, \(r_{H,2}\), should be equal to the radius of some other orbit in different ions. Since \(r_n \propto n^2/Z\), we are looking for ion combinations where the product \(n^2/Z\) remains constant.
03

Comparing with the Second Orbit of He+ ion

For the Helium ion (\(He^+\)), the atomic number is 2. The second orbit will have \(n = 2\), so for the second orbit we have \(n^2/Z = 2^2/2 = 2\). Since for hydrogen \(n = 2\) which gives us \(n^2/Z = 2^2/1 = 4\), these do not match.
04

Comparing with the Third Orbit of Li2+ ion

For the Lithium ion (\(Li^{2+}\)), the atomic number is 3. The third orbit will have \(n = 3\), so we get \(n^2/Z = 3^2/3 = 3\). This does not match the hydrogen value either.
05

Comparing with the Fourth Orbit of He+ ion

Going back to the Helium ion (\(He^+\)) for its fourth orbit when \(n = 4\), we find \(n^2/Z = 4^2/2 = 8\). This also does not match the value of 4 for hydrogen.
06

Comparing with the Fourth Orbit of Be3+ ion

For the Beryllium ion (\(Be^{3+}\)), the atomic number is 4. The fourth orbit will have \(n = 4\), so we obtain \(n^2/Z = 4^2/4 = 4\). This matches the value obtained for hydrogen's second orbit, confirming that the second orbit of an H atom has the same radius as the fourth orbit of a \(Be^{3+}\) ion.

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Key Concepts

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

Bohr Radius
The Bohr radius is a physical constant that represents the most probable distance between the nucleus and the electron in the ground state of a hydrogen-like atom. It is denoted by the symbol a0. The Bohr model, proposed by Niels Bohr in 1913, suggests that electrons move in specific circular orbits or 'shells' around the nucleus with discrete energy levels.

The formula to calculate the radius of an electron's orbit in the hydrogen atom is rn = n2 a0 where n is the principal quantum number and a0 is the Bohr radius, approximately equal to 0.529 × 10-10 meters. This radius plays a key role in determining the size and energy of electronic orbitals in atoms.
Quantum Number
Quantum numbers are sets of numerical values that provide important information about the energy levels and orbitals that electrons occupy in an atom. The principal quantum number, denoted as n, principally determines the energy of the electron in a particular atomic orbital and its distance from the nucleus.

As n increases, the size of the orbital grows, and the electron is located further from the nucleus, which also implies a larger energy orbital. In the context of Bohr's model, n is always a positive integer (1,2,3,...) and directly affects the orbit radius and energy level of an electron.
Electronic Orbitals
Electronic orbitals are regions around an atom's nucleus where electrons are most likely to be found. In the Bohr model, these are visualized as circular paths in which electrons spin at fixed distances from the nucleus. However, the modern quantum mechanical model describes orbitals as three-dimensional shapes with complex geometries like spherical (s orbitals), dumbbell-shaped (p orbitals), and more elaborate shapes for d and f orbitals.

Orbitals are defined by quantum numbers, and changes in energy levels correspond to electron transitions between these orbitals. Each orbital can hold a maximum of two electrons, with their spins oriented in opposite directions.
Hydrogen Atom
The hydrogen atom is the simplest atom and a fundamental component of chemistry and physics. It consists of a single proton within its nucleus and one electron bound to the nucleus by electromagnetic force.

The energy levels of a hydrogen atom are quantized, meaning the electron can only exist in certain specific orbits with predetermined energies. When an electron jumps between these energy levels, the atom absorbs or emits light at specific wavelengths. This behavior is crucial in understanding atomic spectra and the Bohr model's treatment of the hydrogen atom as the cornerstone for explaining its spectral lines.
Ion Energy Levels
Ion energy levels pertain to the discrete energies that electrons can have in an ion, which is an atom with a net electric charge due to the loss or gain of one or more electrons. The energy levels in ions are affected by the nuclear charge and the electron-electron repulsions within the ion.

For hydrogen-like ions, where only one electron is present, the energy levels can be determined using a modified version of the Bohr formula: En = -Z2 E0 / n2, where E0 is the ground state energy of the hydrogen atom, Z is the atomic number, and n is the principal quantum number. A higher nuclear charge (Z) results in more tightly bound electrons, and thus higher ionization energies.

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Most popular questions from this chapter

A ruby laser produces radiations of wavelength \(662.6 \mathrm{~nm}\) in pulses whose duration are \(1.0 \times 10^{-9} \mathrm{~s}\). If the laser produces \(0.36 \mathrm{~J}\) of energy per pulse, how many photons are produced in each pulse? (a) \(1.2 \times 10^{9}\) (b) \(1.2 \times 10^{27}\) (c) \(1.2 \times 10^{18}\) (d) \(1.2 \times 10^{15}\)

Photoelectric emission is observed from a metal surface for frequencies \(v_{1}\) and \(v_{2}\) of the incident radiation \(\left(v_{1}>v_{2}\right)\). If maximum kinetic energies of the photoelectrons in the two cases are in the ratio \(1: K\), then the threshold frequency for the metal is given by (a) \(\frac{v_{2}-v_{1}}{K-1}\) (b) \(\frac{K v_{2}-v_{1}}{K-1}\) (c) \(\frac{K v_{1}-v_{2}}{K}\) (d) \(\frac{K v_{1}-v_{2}}{K-1}\)

Imagine an atom made up of a stationary proton and a hypothetical particle of double the mass of electron but having the same charge as the electron. Apply Bohr atomic model and consider all possible transitions of this hypothetical particle directly to the first excited state. The longest wavelength photon that will be emitted has wavelength (given in terms of Rydberg constant \(R\) for the hydrogen atom) equal to (a) \(\frac{9}{5 R}\) (b) \(\frac{36}{5 R}\) (c) \(\frac{18}{5 R}\) (d) \(\frac{4}{R}\)

The electronic structure of zinc \((Z=30)\) is \(2,8,18,2 .\) The electronic structure of gallium \((Z=31)\) will be (a) \(2,8,18,2,1\) (b) \(2,8,19,2\) (c) \(2,8,18,3\) (d) \(2,8,19,3\)

Which of the following element will have same number of electrons in s-as well as p-type of orbitals? (a) \(\mathrm{Fe}(Z=26)\) (b) \(\operatorname{Mg}(Z=12)\) (c) \(\operatorname{Ne}(Z=10)\) (d) \(\operatorname{Ar}(Z=18)\)

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