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A hydrogen atom is in a state with energy -1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Short Answer

Expert verified
The angular momentum is \( 3\hbar \).

Step by step solution

01

Understanding the Energy Level

In the Bohr model, the energy levels of a hydrogen atom are given by the formula: \[ E_n = \frac{-13.6 \, \text{eV}}{n^2} \] where \( E_n \) is the energy of the electron at a particular level \( n \). We know that the energy is \(-1.51 \, \text{eV}\). We need to find the principal quantum number \( n \) using this formula.
02

Calculate the Principal Quantum Number

Set up the equation based on the known energy level:\[ \frac{-13.6 \, \text{eV}}{n^2} = -1.51 \, \text{eV} \]Solve for \( n^2 \) by dividing both sides by \(-13.6 \, \text{eV}\):\[ n^2 = \frac{13.6}{1.51} \approx 9 \]Taking the square root of both sides, we find:\[ n = 3 \] Thus, the electron is in the third energy level.
03

Find the Angular Momentum

In the Bohr model, the angular momentum \( L \) of an electron is quantized and given by:\[ L = n \hbar \]where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant, and \( n \) is the principal quantum number. Since we found \( n = 3 \), the angular momentum is:\[ L = 3 \hbar \]

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Key Concepts

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

Hydrogen Atom
The hydrogen atom, the simplest of all atoms, consists of a single electron orbiting a single proton. This simplicity makes it an excellent starting point for understanding atomic structures. In a hydrogen atom, the electron moves around the nucleus, forming various energy levels. These levels are dictated by specific rules as per Bohr's model. According to Bohr, electrons would only occupy certain orbits, called "quantized orbits," without losing energy by radiation. This concept was revolutionary because it explained the hydrogen spectrum observed.

The hydrogen atom's structure allows us to study fundamental principles in quantum mechanics. Since hydrogen holds only one electron, the interactions are minimal, providing a clean situation to interpret spectral lines. Hydrogen's spectrum consists of distinct lines, which correspond to various transitions of the electron between these energy levels.

Understanding the hydrogen atom sets the foundation for all other atomic structures, as more complex atoms can be interpreted as variations on this simplest case.
Angular Momentum
In the Bohr model, angular momentum refers to the motion of an electron around the nucleus. This concept specifies that the angular momentum of the electron is quantized, meaning it can only take specific values. The idea of quantization was pivotal as it provided a theoretical explanation for the distinct energy levels in an atom.

A key equation that represents angular momentum in Bohr's model is:
  • Angular momentum, \( L = n \hbar \)
where \( n \) is the principal quantum number and \( \hbar = \frac{h}{2\pi} \), known as the reduced Planck's constant. In our example, if the principal quantum number is 3, the angular momentum of the electron is \( 3 \hbar \).

The concept of angular momentum quantization was groundbreaking as it explained the stability of electrons in atoms, supporting why they didn't spiral into the nucleus. Instead, electrons remain in stable orbits or shells.
Principal Quantum Number
The principal quantum number, denoted as \( n \), is a fundamental element in quantum mechanics. It describes the size and energy of the orbital, which tells us the distance of the electron from the nucleus. In the Bohr model, the principal quantum number is a crucial factor in determining the energy levels of an atom.

In the exercise, we used the formula for energy levels in the hydrogen atom:
  • \( E_n = \frac{-13.6 \, \text{eV}}{n^2} \)

We set the known energy \(-1.51 \, \text{eV}\) into the formula to solve for \( n \), resulting in \( n = 3 \). This tells us the electron resides on the third energy level or orbit. Each value of \( n \) corresponds to a major energy level and is sometimes referred to as a shell.

Understanding the principal quantum number is key to knowing the distribution of electrons in an atom. It assists in explaining atomic behavior and spectra.
Energy Levels
Energy levels are core to the structure of atoms. In the Bohr model, electrons reside in specific energy levels or orbits that determine their potential energy within an atom. These energy levels are quantized, meaning electrons can jump between these levels but cannot exist in between them.

For a hydrogen atom, the energy levels can be determined by:
  • \( E_n = \frac{-13.6 \, \text{eV}}{n^2} \)
where \( E_n \) represents the energy of a level, and \( n \) is the principal quantum number. Energy levels become less negative with increasing \( n \), indicating less tightly bound electrons as they move to higher energy states.

Each transition of an electron between levels emits or absorbs a photon, explaining the spectral lines seen in hydrogen's emission and absorption spectra. The quantization of these energy levels highlights the quantum nature of atomic structures.

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Most popular questions from this chapter

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (\(Hint\): In each case the initial kinetic energy is negligible.)

(a) What is the energy of a photon that has wavelength 0.10 \(\mu\)m ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 \(\mu\)m in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

A beam of electrons is accelerated from rest through a potential difference of 0.100 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at \(\pm\)14.6\(^\circ\) from the original direction of the beam. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K/E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K/E\) for a hydrogen atom (mass 1.67 \(\times\) 10\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

The radii of atomic nuclei are of the order of 5.0 \(\times\) 10\(^{-15}\) m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 \(\times\) 10\(^{-15}\) m. On the basis of your result, could there be electrons within the nucleus? (\(Note\): It is interesting to compare this result to that of Problem 39.72.)

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