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What is the energy of the orbiting electron in a hydrogen atom with a quantum number of \(45 ?\)

Short Answer

Expert verified
Answer: The energy of the orbiting electron in a hydrogen atom with a quantum number of 45 is approximately -0.0067 eV.

Step by step solution

01

Write down the given information and formula

The quantum number of the hydrogen atom is given: \(n=45\). The formula for the energy levels of the hydrogen atom is: \(E_n = -\frac{13.6 \text{eV}}{n^2}\)
02

Substitute the given quantum number in the formula

We need to find the energy of the orbiting electron in the hydrogen atom. Substitute the given quantum number \(n=45\) in the energy formula: \(E_{45} = -\frac{13.6 \text{eV}}{45^2}\)
03

Calculate the energy of the electron

Now, calculate the energy of the electron in the hydrogen atom with the given quantum number: \(E_{45} = -\frac{13.6 \text{eV}}{45^2} = -\frac{13.6 \text{eV}}{2025} = -0.006716049 \text{eV}\) The energy of the orbiting electron in a hydrogen atom with a quantum number of \(45\) is approximately \(-0.0067 \text{eV}\).

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Key Concepts

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

Quantum Number
Quantum numbers are essential in describing the specific energy levels within an atom. They are like an address system for electrons, indicating where an electron might be found in an atom and its energy level.
For any atom, the principal quantum number, denoted as \( n \), primarily determines the energy level. It takes positive integer values like 1, 2, 3, and so on.
  • A higher quantum number means that the electron is in a larger orbit, farther from the nucleus, and usually at a higher energy level.
  • In the given exercise, the quantum number is 45, which means the electron is quite far from the nucleus and in a very high energy orbit. Despite its large value, the actual energy calculated is small, as observed.
Understanding quantum numbers helps in predicting the arrangement of electrons in atoms and thus is crucial in quantum mechanics.
Bohr Model
The Bohr Model of the atom was a pivotal development in the understanding of atomic structure. Proposed by Niels Bohr in 1913, it introduced the concept of quantized orbits for electrons.
This model suggests that electrons move in fixed circular paths, called orbits, around the nucleus, with each orbit corresponding to a specific energy level.
  • In Bohr's model, electrons transition between orbits by absorbing or emitting energy in discrete quanta.
  • The specific formula used in the exercise, \( E_n = -\frac{13.6 \text{eV}}{n^2} \), comes from Bohr's work and predicts the energy levels of the hydrogen atom. It remains useful today for educational purposes, although it has been refined by newer quantum mechanical models.
The model is crucial in illustrating the quantization of energy levels and explaining phenomena such as atomic spectra.
Electron Energy Calculation
Calculating electron energy in atoms involves specific formulas derived from models like Bohr's. In the hydrogen atom, this calculation is straightforward due to its simplicity as a single-electron system.
Using the formula \( E_n = -\frac{13.6 \text{eV}}{n^2} \), we can determine the energy of an orbiting electron if the quantum number \( n \) is known.
  • This equation shows that the energy is inversely proportional to the square of the quantum number, \( n^2 \). Thus, as \( n \) increases, the energy becomes less negative, approaching zero.
  • For the quantum number 45, as calculated, the energy is \(-0.0067 \text{eV}\), indicating a weakly bound state slightly negative, implying an electron loosely associated with the nucleus at this high level.
Understanding electron energy calculations helps in analyzing atomic behavior and electronic transitions comprehensively, especially in hydrogen-like systems.

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Most popular questions from this chapter

A hydrogen atom is in its fifth excited state, with principal quantum number \(n=6 .\) The atom emits a photon with a wavelength of \(410 \mathrm{nm}\). Determine the maximum possible orbital angular momentum of the electron after emission.

The hydrogen atom wave function \(\psi_{200}\) is zero when \(r=2 a_{0} .\) Does this mean that the electron in that state can never be observed at a distance of \(2 a_{0}\) from the nucleus or that the electron can never be observed passing through the spherical surface defined by \(r=2 a_{0}\) ? Is there a difference between those two descriptions?

The muon has the same charge as an electron but a mass that is 207 times greater. The negatively charged muon can bind to a proton to form a new type of hydrogen atom. How does the binding energy \(E_{\mathrm{B} \mu}\) of the muon in the ground state of a muonic hydrogen atom compare with the binding energy \(E_{\mathrm{Be}}\) of an electron in the ground state of a conventional hydrogen atom? a) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right|\) d) \(\left|E_{\mathrm{B} \mu}\right| \approx 200 \mid E_{\mathrm{Be}}\) b) \(\left|E_{\mathrm{B} \mu}\right| \approx 100\left|E_{\mathrm{Be}}\right|\) e) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 200\) c) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 100\)

Consider a muonic hydrogen atom, in which an electron is replaced by a muon of mass \(105.66 \mathrm{MeV} / \mathrm{c}^{2}\) that orbits a proton. What are the first three energy levels of the muon in this type of atom?

By what percentage is the electron mass changed in using the reduced mass for the hydrogen atom? What would the reduced mass be if the proton had the same mass as the electron?

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