A Group Of Hydrogen Atoms In A Discharge Tube Emit Violet Light Of Wavelength 410 Nm. Determine The Quantum
Understanding the behavior of hydrogen atoms and their emission spectra is fundamental in atomic physics and quantum mechanics. When hydrogen atoms are excited in a discharge tube, they emit light at specific wavelengths, revealing vital information about their energy levels and the quantized nature of atomic energy states. This article delves into the phenomenon of hydrogen emission lines, focusing on the violet light with a wavelength of 410 nm, and guides you through the process of determining the quantum involved in this emission.
Introduction to Atomic Spectra and Hydrogen Emission Lines
Hydrogen, the simplest atom consisting of one proton and one electron, exhibits a distinctive emission spectrum when its electrons transition between energy levels. These spectral lines are characteristic signatures that allow scientists to analyze atomic structures and validate quantum theories.
What Are Atomic Emission Spectra?
Atomic emission spectra arise when electrons in an atom transition from higher to lower energy states, releasing photons with energies corresponding to the difference between these levels. These photons manifest as specific wavelengths of light, forming a spectrum unique to each element.
Hydrogen’s Spectral Series
Hydrogen's emission spectrum comprises several series, each corresponding to electron transitions ending at a particular energy level:
- Lyman Series: Transitions ending at the n=1 level (ultraviolet region)
- Balmer Series: Transitions ending at n=2 (visible region)
- Paschen Series: Transitions ending at n=3 (infrared region)
- And others such as Brackett and Pfund series
The violet light at 410 nm falls within the visible region, generally associated with the Balmer series, which involves transitions ending at n=2.
Understanding the Quantum Number and Transition in Hydrogen
In atomic physics, the energy levels of electrons in hydrogen are quantified by the principal quantum number, n, which takes positive integer values (n=1, 2, 3, ...). When an electron transitions between two energy levels, the change in the quantum number determines the emitted photon’s energy and wavelength.
Energy Levels and Bohr Model of Hydrogen
The Bohr model provides a simplified way to calculate the energy of electrons in hydrogen:
\[ E_n = -\frac{13.6\, \text{eV}}{n^2} \]
where:
- \(E_n\) is the energy of level n
- 13.6 eV is the ionization energy of hydrogen
The negative sign indicates that the electron is bound within the atom. When an electron transitions from a higher level \(ni\) to a lower level \(nf\), the energy of the emitted photon is:
\[ \Delta E = E{nf} - E{ni} \]
which corresponds to the photon energy:
\[ E_{photon} = h \nu = \frac{hc}{\lambda} \]
where:
- \(h\) is Planck’s constant (\(6.626 \times 10^{-34}\, \text{Js}\))
- \(c\) is the speed of light (\(3.0 \times 10^8\, \text{m/s}\))
- \(\lambda\) is the wavelength of emitted light
Calculating the Quantum Transition for the 410 nm Violet Emission
Given the wavelength of the emitted violet light, the goal is to determine the quantum transition responsible for this emission—specifically, the initial and final energy levels involved.
Step 1: Convert Wavelength to Energy
Using the relation:
\[ E_{photon} = \frac{hc}{\lambda} \]
where:
- \(\lambda = 410\, \text{nm} = 410 \times 10^{-9}\, \text{m}\)
Calculating:
\[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{410 \times 10^{-9}} \]
\[ E_{photon} \approx \frac{1.9878 \times 10^{-25}}{4.10 \times 10^{-7}} \]
\[ E_{photon} \approx 4.846 \times 10^{-19}\, \text{J} \]
Converting Joules to electronvolts (1 eV = \(1.602 \times 10^{-19}\) J):
\[ E_{photon} \approx \frac{4.846 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.027\, \text{eV} \]
Step 2: Determine the Transition Level
Since the emission occurs in the visible region and aligns with the Balmer series, which involves transitions ending at level \(n=2\), the initial level \(n_i\) is higher than 2.
Using the energy level formula:
\[ E_n = -\frac{13.6\, \text{eV}}{n^2} \]
The energy difference:
\[ \Delta E = E{nf} - E{ni} \]
Given \(n_f = 2\):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4\, \text{eV} \]
So,
\[ \Delta E = E{ni} - (-3.4\, \text{eV}) = E{ni} + 3.4\, \text{eV} \]
But since the photon energy is approximately 3.027 eV, and the transition is from \(n_i\) to 2:
\[ E{ni} + 3.4\, \text{eV} \approx 3.027\, \text{eV} \]
Rearranged:
\[ E{ni} \approx 3.027 - 3.4 = -0.373\, \text{eV} \]
Now, equate this to the energy level formula:
\[ E{ni} = -\frac{13.6}{n_i^2} \]
which gives:
\[ -\frac{13.6}{n_i^2} \approx -0.373 \]
Simplify:
\[ \frac{13.6}{n_i^2} \approx 0.373 \]
\[ n_i^2 \approx \frac{13.6}{0.373} \]
\[ n_i^2 \approx 36.44 \]
\[ n_i \approx \sqrt{36.44} \approx 6.04 \]
Since \(n_i\) must be an integer, the closest integer is 6.
Result: The transition responsible for the violet emission at 410 nm is from \(n=6\) to \(n=2\).
Summary of the Quantum Transition
| Initial Level (\(ni\)) | Final Level (\(nf\)) | Wavelength (nm) | Transition Type |
|-------------------------|----------------------|-----------------|------------------|
| 6 | 2 | 410 | Balmer Series |
This transition from \(n=6\) to \(n=2\) corresponds to the violet emission observed.
Significance of the Quantum Transition in Atomic Physics
Understanding such quantum transitions is pivotal in validating quantum theory and the Bohr model's accuracy. The quantized energy levels explain why hydrogen emits light at specific wavelengths, leading to the formation of the characteristic emission lines. Analyzing the transition from \(n=6\) to \(n=2\) confirms the energy quantization and supports the concept of discrete energy states.
Applications in Modern Science and Technology
- Spectroscopy: Used to identify elements in stars and distant celestial bodies.
- Laser Technology: Understanding atomic transitions underpins the development of laser sources.
- Quantum Mechanics: Empirical validation of the quantized energy levels in atoms.
- Astrophysics: Analyzing spectral lines to determine the composition and physical conditions of astronomical objects.
Conclusion
The violet light emitted by hydrogen atoms in a discharge tube with a wavelength of 410 nm results from a specific electronic transition between energy levels. Calculations based on quantum physics reveal that this emission corresponds to a transition from the \(n=6\) level to the \(n=2\) level. This process exemplifies the discrete nature of atomic energy levels and underscores the importance of quantum mechanics in explaining atomic spectra.
By understanding and calculating these quantum transitions, scientists can interpret spectral data, explore atomic structures, and develop technological applications that rely on precise control of atomic emissions. The study of hydrogen's emission lines continues to be a cornerstone in the field of atomic physics and quantum mechanics, illustrating the profound link between microscopic quantum phenomena and observable spectral features.
References
- Griffiths, D. (2017). Introduction to Quantum Mechanics. Cambridge University Press.
- Tipler, P. A., & Llewellyn, R. (2013).
