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Chapter 27: Problem 53

In the Bohr model of the hydrogen atom, let \(R, V\) and \(E\) represent the radius of the orbit, the speed of electron and the total energy of the electron respectively. Which of the following quantities is proportional to quantum number \(n ?\) (a) \(\frac{R}{E}\) (b) \(\frac{E}{V}\) (c) \(R \bar{E}\) (d) \(V \underline{R}\)

Short Answer

Expert verified
(d) \( V \underline{R} \) is proportional to \( n \).

Step by step solution

01

Understanding the Bohr Model

In the Bohr model, the radius of the orbit is given by \( R_n = n^2 a_0 \), where \( a_0 \) is the Bohr radius and \( n \) is the principal quantum number. The velocity \( V \) of the electron is \( V_n = \frac{k}{n} \), where \( k \) is a constant. The energy \( E \) of the electron is given by \( E_n = \frac{-E_0}{n^2} \), where \( E_0 \) is a constant.
02

Evaluating \( \frac{R}{E} \)

Calculate \( \frac{R_n}{E_n} = \frac{n^2 a_0}{-\frac{E_0}{n^2}} = -\frac{n^4 a_0}{E_0} \). Clearly, \( \frac{R}{E} \) is proportional to \( n^4 \).
03

Evaluating \( \frac{E}{V} \)

For \( \frac{E}{V} \), substitute the known relationships: \( \frac{E_n}{V_n} = \frac{-\frac{E_0}{n^2}}{\frac{k}{n}} = -\frac{E_0 n}{k n^2} = -\frac{E_0}{k n} \), showing this is proportional to \( \frac{1}{n} \).
04

Evaluating \( R \bar{E} \)

Calculate \( R_n E_n = n^2 a_0 \times \left(-\frac{E_0}{n^2}\right) = -a_0 E_0 \). There is no \( n \) dependence, thus it is not proportional to \( n \).
05

Evaluating \( V \underline{R} \)

Calculate \( V_n R_n = \frac{k}{n} \times n^2 a_0 = k n a_0 \). Clearly, \( V \underline{R} \) is proportional to \( n \).
06

Concluding with the Correct Choice

By analyzing each option, we find that only \( V \underline{R} \) (option d) is directly proportional to \( n \).

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Key Concepts

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

Quantum Number
In the Bohr model, quantum numbers play a crucial role in determining the properties of atomic orbits. The principal quantum number, denoted by \( n \), is a positive integer. It is one of the simplest quantum numbers and is responsible for quantifying the size and energy level of the electron's orbit in an atom.
  • Higher values of \( n \) indicate that the electron is in a higher energy state and occupies a larger orbit.
  • As \( n \) increases, the electron becomes less tightly bound to the nucleus, making it easier to ionize.
This quantum number impacts other properties such as the radius of the orbit and energy levels, as we'll see in the context of the hydrogen atom.
Hydrogen Atom
The hydrogen atom, being the simplest atom, consists of a single proton and a single electron. It's typically used to introduce atomic and quantum physics concepts due to its simplicity.
In the Bohr model of the hydrogen atom:
  • The electron revolves around the nucleus in circular orbits, similar to planets orbiting a star.
  • The energies and radii of these orbits are quantized, meaning they can only take on specific values determined by the quantum number \( n \).
This model laid the groundwork for quantum mechanics, even though the understanding of electron orbits has since evolved to a more complex description using wave functions. Nonetheless, the Bohr model remains a valuable tool in understanding fundamental concepts like electron orbit radius and quantized energy levels.
Electron Orbit Radius
In the Bohr model, the radius of an electron's orbit is directly linked to the principal quantum number \( n \). The formula for the radius \( R_n \) of the nth orbit is \[ R_n = n^2 a_0 \] where \( a_0 \) is the Bohr radius (approximately \( 0.529 \, \text{脜} \)).
  • This equation tells us that the orbit's radius increases with the square of \( n \).
  • An electron in a higher energy level \( n \) has a larger orbit radius than one in a lower energy level.
  • Larger orbits mean the electron is further from the nucleus, resulting in lower binding energy.
This quantized relationship between \( n \) and the radius is a pivotal aspect of understanding atomic structure in elemental hydrogen and beyond.
Electron Speed
The speed of an electron in a hydrogen atom, according to the Bohr model, is inversely related to the principal quantum number. The electron's velocity \( V_n \) in a given orbit is given by:\[ V_n = \frac{k}{n} \] where \( k \) is a constant derived from fundamental constants.
  • This means that as \( n \) increases, the speed of the electron decreases.
  • At higher energy levels, the electron moves more slowly around the nucleus.
  • Slower electron speeds are associated with larger orbit radii and higher energy levels.
Understanding this inverse relationship helps explain and predict the behavior of electrons in different atomic energy levels, particularly their speeds and kinetic energies.
Electron Energy
In the context of the Bohr model for the hydrogen atom, the total energy of an electron in its orbit is another property dictated by the principal quantum number \( n \). The electron's energy \( E_n \) is expressed by the formula:\[ E_n = \frac{-E_0}{n^2} \] where \( E_0 \) is a constant representing the ground state energy.
  • This shows that the energy becomes less negative with increasing \( n \), meaning the electron has higher energy at higher orbits.
  • The energy levels are inversely proportional to the square of the quantum number, which reflects the quantized nature of an electron's energy states.
  • As \( n \) increases, the electron becomes less tightly bound, ultimately leading to ionization if energy is sufficiently increased.
This understanding of electron energy levels is fundamental in atomic physics, lending insight into why electrons occupy specific energy states in atoms.

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Most popular questions from this chapter

An X-ray tube with copper target emit \(K_{\alpha}\) line of wavelength \(1.50 \AA\). What should be the minimum voltage through which electrons one to be accelerated to produce this wavelength of X-rays. \(\left(h=6.6 \times 10^{-34} \mathrm{Js}, c=3 \times 10^{8} \mathrm{~ms}^{-1}\right)\) (a) \(82.8 \mathrm{~V}\) (b) \(8280 \mathrm{~V}\) (c) \(82801 \mathrm{~V}\) (d) \(828 \mathrm{~V}\)

Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into \(p+\bar{e}+\bar{v}\). If one of the neutrons in Triton decays, it would transform into \(\mathrm{He}^{3}\) nucleus. This does not happen. This is because [NCERT Exemplar] (a) Triton energy is less than that of a \(\mathrm{He}^{3}\) nucleus. (b) The electron created in the beta decay process cannot remain in the nucleus. (c) both the neutrons in trition have to decay simultaneously resulting in a nucleus with 3 protons, which is not a \(\mathrm{He}^{3}\) nucleus. (d) because free neutrons decay due to external perturbations which is absent in a triton nucleus.

An oil drop carrying a charge \(q\) has a mass \(m \mathrm{~kg}\). It is falling freely in air with terminal speed \(v\). The electric field required to make, the drop move upwards with the same speed is (a) \(\frac{m g}{q}\) (b) \(\frac{2 m g}{q}\) (c) \(\frac{m g v}{q^{2}}\) (d) \(\frac{2 m g v}{q}\)

\(\mathrm{O}_{2}\) molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms. (a) is not important because nuclear forces are short-ranged (b) is as important as electrostatic force for binding the two atoms. (c) cancels the repulsive electrostatic force between the nuclei (d) is not important because oxygen nucleus have equal number of neutrons and protons.

Two nucleons are at a separation of \(1 \mathrm{fm}\). The net force between them is \(F_{1}\), if both neutrons, \(F_{2}\) if both are protons and \(F_{3}\) if one is a proton and the other is a neutron. (a) \(F_{1}>F_{2}>F_{3}\) (b) \(F_{2}>F_{1}>F_{3}\) (c) \(F_{1}=F_{3}>F_{1}\) (d) \(F_{1}=F_{2}>F_{3}\)
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