Problem 13
Question
What is \(\Delta E\) for the transition of an electron from \(n=6\) to \(n=3\) in a Bohr hydrogen atom? What is the frequency of the spectral line produced?
Step-by-Step Solution
Verified Answer
The energy change \(\Delta E\) for the transition of an electron from \(n=6\) to \(n=3\) in a Bohr hydrogen atom is -1.133 eV or -1.815x10^-19 Joules. The frequency of the spectral line produced is 2.74 x 10^{14} Hz.
1Step 1: Calculate Energy at Level 6
First, calculate the energy of the electron at energy level \(n=6\) using the formula \(E_n = -13.6/ n^2\). Here, \(n = 6\), so \(E_{n=6} = -13.6/6^2 = -0.3778\) eV
2Step 2: Calculate Energy at Level 3
Next, calculate the energy of the electron at energy level \(n=3\) using the same formula. Here, \(n = 3\). Hence, \(E_{n=3} = -13.6/3^2 = -1.511\) eV
3Step 3: Calculate ∆E
Now calculate the change in energy, \(\Delta E \), which is the energy difference between the two levels. \(\Delta E = E_{n=3} - E_{n=6} = -1.511 - (-0.3778) = -1.133\) eV
4Step 4: Convert Energy to Joules
We will convert the energy difference from electron Volts (eV) to Joules (J) to utilize it further: 1eV = \(1.602\times10^{-19}\) J. Thus \(\Delta E = -1.133 \times 1.602\times10^{-19} = -1.815\times10^{-19}\) Joules.
5Step 5: Calculate Frequency
Now, calculate the frequency of the spectral line produced using the equation \(E = hf\), where \(h\) is Planck's constant. Solving for \(f\), we get \(f = \Delta E / h\). Given \(h = 6.626\times10^{-34}\) Js, \(f = -1.815\times10^{-19} / 6.626\times10^{-34} = 2.74 \times 10^{14}\) Hz.
Key Concepts
Energy LevelsHydrogen AtomSpectral Lines
Energy Levels
In the Bohr model of the hydrogen atom, electrons occupy specific orbits around the nucleus, which are called energy levels. Each level is denoted by an integer, known as the principal quantum number, symbolized as \(n\). The energy of any electron in these levels can be calculated using the formula:
As a result, electrons closer to the nucleus (lower \(n\) values) have more negative energy, indicating they're more tightly bound. Energizing an electron moves it to a higher level (higher \(n\) value), making its energy less negative.
When electrons transition between these energy levels, they absorb or emit specific amounts of energy, corresponding to differences between the energy levels involved.
- \(E_n = -\frac{13.6}{n^2}\) (eV)
As a result, electrons closer to the nucleus (lower \(n\) values) have more negative energy, indicating they're more tightly bound. Energizing an electron moves it to a higher level (higher \(n\) value), making its energy less negative.
When electrons transition between these energy levels, they absorb or emit specific amounts of energy, corresponding to differences between the energy levels involved.
Hydrogen Atom
The hydrogen atom is the simplest atom, consisting of a single proton as its nucleus and one orbiting electron. It's often used to explain atomic physics due to its simplicity and serve as an ideal example for models like Bohr's. In Bohr's model, the electron orbits the nucleus in circular paths with quantized energies. These quantization rules set by Bohr state:
This simple structure makes the hydrogen atom perfect for studying fundamental principles of quantum mechanics, as calculations and observations are simpler compared to other elements.
- Electrons exist only in certain orbits with defined energies.
- The movement between orbits involves absorbing or releasing energy.
This simple structure makes the hydrogen atom perfect for studying fundamental principles of quantum mechanics, as calculations and observations are simpler compared to other elements.
Spectral Lines
When an electron transitions between energy levels in a hydrogen atom, it emits or absorbs radiation of specific frequencies, which are observed as spectral lines. Each possible transition corresponds to a particular frequency or wavelength of light. This emission or absorption can be analyzed using spectral lines.
Spectral lines serve as a fingerprint for elements:
This is part of the Balmer series, specific to hydrogen, which explains visible spectral lines in hydrogen's light emission spectrum. This model helps astronomers and physicists identify elements in stars and other celestial bodies.
Spectral lines serve as a fingerprint for elements:
- Each element's lines are unique due to their unique energy levels.
- They can be seen in absorption or emission spectra.
This is part of the Balmer series, specific to hydrogen, which explains visible spectral lines in hydrogen's light emission spectrum. This model helps astronomers and physicists identify elements in stars and other celestial bodies.
Other exercises in this chapter
Problem 9
Use the Balmer equation (8.2) to determine (a) the frequency, in \(s^{-1}\), of the radiation corresponding to \(n=5\) (b) the wavelength, in nanometers, of the
View solution Problem 11
Use Planck's equation (8.3) to determine (a) the energy, in joules per photon, of radiation of frequency \(7.39 \times 10^{15} \mathrm{s}^{-1}\) (b) the energy,
View solution Problem 14
What is \(\Delta E\) for the transition of an electron from \(n=5\) to \(n=2\) in a Bohr hydrogen atom? What is the frequency of the spectral line produced?
View solution Problem 16
The Lyman series of the hydrogen spectrum can be represented by the equation $$\nu=3.2881 \times 10^{15} \mathrm{s}^{-1}\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\ri
View solution