U4AOS1 Topic 25: Emission Spectra

Emission spectra are a fundamental concept in physics and chemistry, providing crucial insights into the nature of atoms and molecules. They occur when atoms or molecules absorb energy and subsequently release it in the form of light, which can be analyzed to identify the elements or compounds present.

What is Emission Spectra?

Emission spectra are the range of wavelengths emitted by atoms or molecules when they transition from a higher energy state to a lower energy state. These spectra are unique to each element or compound, acting as a "fingerprint" that can be used to identify substances and study their properties.

The Process of Emission

  1. Energy Absorption:

    • When an atom or molecule absorbs energy, usually from heat or electromagnetic radiation, its electrons are excited to a higher energy level or orbital. This is an unstable state, and the system will seek to return to a more stable, lower energy state.

  2. Relaxation and Photon Emission:

    • As the excited electrons return to their original, lower energy state, the excess energy is released in the form of photons (light). The energy of the emitted photons corresponds to the difference in energy levels between the excited and ground states.

  3. Emission Spectrum:

    • The emitted light can be dispersed using a prism or diffraction grating to produce a spectrum. This spectrum typically consists of distinct lines (in the case of atomic spectra) or bands (for molecular spectra) at specific wavelengths. These lines or bands are characteristic of the element or molecule.

Types of Emission Spectra

  1. Atomic Emission Spectrum:

    • This spectrum is produced by atoms and consists of a series of discrete lines. Each line corresponds to a specific transition between energy levels of the electrons in the atom. The wavelengths of these lines are unique to each element. For example, hydrogen's emission spectrum includes lines known as the Balmer series in the visible range.

  2. Molecular Emission Spectrum:

    • Molecules can emit light at multiple wavelengths due to electronic, vibrational, and rotational transitions. This type of spectrum often appears as bands rather than discrete lines, covering a broader range of wavelengths. The emission spectrum of molecules is more complex than that of atoms.

Applications of Emission Spectra

  1. Chemical Analysis:

    • Emission spectra are used in spectroscopy to identify elements and compounds. By analyzing the emitted light from a sample, scientists can determine its composition. This technique is widely used in analytical chemistry, astronomy, and environmental monitoring.

  2. Astronomy:

    • Emission spectra are crucial for understanding the composition of stars and galaxies. By studying the light emitted by celestial objects, astronomers can determine their chemical composition, temperature, density, mass, distance, luminosity, and relative motion.

  3. Material Science:

    • Emission spectra help in characterizing materials, studying their properties, and developing new materials with specific properties. This is important in fields like semiconductors and nanotechnology.

  4. Medical Diagnostics:

    • Techniques such as fluorescence spectroscopy use emission spectra to detect and analyze biological molecules, aiding in medical diagnostics and research.

Example 1

An electron in a hydrogen atom transitions from the n=4n = 4 energy level to the n=2n = 2 energy level. Calculate the wavelength of the emitted photon.

The wavelength of the emitted photon can be calculated using the Rydberg formula for hydrogen:
1λ=RH(1n12−1n22)\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
where:
  • λ\lambda is the wavelength,
  • RHR_H is the Rydberg constant (1.097×107 m−11.097 \times 10^7\ \text{m}^{-1}),
  • n1n_1 and n2n_2 are the principal quantum numbers of the two energy levels (with n2>n1n_2 > n_1).
For the transition from n=4n = 4 to n=2n = 2:
1λ=1.097×107(122−142)\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
1λ=1.097×107(14−116)\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)
1λ=1.097×107(4−116)\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{4 - 1}{16} \right)
1λ=1.097×107×316\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16}
1λ=2.06×106 m−1\frac{1}{\lambda} = 2.06 \times 10^6\ \text{m}^{-1}
λ=12.06×106\lambda = \frac{1}{2.06 \times 10^6}
λ≈4.85×10−7 m\lambda \approx 4.85 \times 10^{-7}\ \text{m}
So, the wavelength of the emitted photon is approximately 485 nm485\ \text{nm}.

Example 2

A spectral line in the emission spectrum of an element has a wavelength of 600 nm600\ \text{nm}. Calculate the frequency of this spectral line.

The frequency ν\nu of the emitted photon can be calculated using the relationship between wavelength λ\lambda and frequency:
ν=cλ\nu = \frac{c}{\lambda}
where:
  • cc is the speed of light (3.00×108 m/s3.00 \times 10^8\ \text{m/s}),
  • λ\lambda is the wavelength (600 nm=600×10−9 m600\ \text{nm} = 600 \times 10^{-9}\ \text{m}).
Substitute the values:
ν=3.00×108600×10−9\nu = \frac{3.00 \times 10^8}{600 \times 10^{-9}}
ν=5.00×1014 Hz\nu = 5.00 \times 10^{14}\ \text{Hz}
So, the frequency of the spectral line is 5.00×1014 Hz5.00 \times 10^{14}\ \text{Hz}.

Exercise 1

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