Helium (He) Gas discharge tube: Bright line emission spectra viewed through a spectroscope: with Wavelength 400 nm 450 nm 500 nm 550 nm 600 nm 660 nm 700 nm (4pts) Attach your drawing of the helium lamp spectrum here. Neon (Ne) Gas discharge tube: Bright line emission spectra viewed through a spectroscope with Wavelength 650 nm 700 nm 400 nm 450 nm 500 nm 550 nm 600 nm Bright Line Spectrum (4pts) Attach your drawing of the neon lamp spectrum here. Krypton (Kr) Gas discharge tube: Bright line emission spectra viewed through a spectroscope: Wavelength 450 nm 500 nm 550 nm 600 nm 650 nm 700 nm 400 nm (4pts) Attach your drawing of the krypton lamp spectrum here. Calculate the theoretical energy value of each wavelength of light in the helium spectrum to complete the following chart. Helium gas emission spectrum Theoretical energy (J) Wavelength (nm) Color Wavelength 700 nm 650 nm 400 nm 450 nm 500 nm 550 nm 600 nm Bright Line Spectrum (4pts) Attach your drawing of the krypton lamp spectrum here. Calculate the theoretical energy value of each wavelength of light in the helium spectrum to complete the following chart. Helium gas emission spectrum Theoretical energy (J) Wavelength (nm) Color 447 violet 471 blue 501 green 587 yellow 668 red
The Correct Answer and Explanation is:
Helium Gas Emission Spectrum: Theoretical Energy Calculation and Explanation
To calculate the theoretical energy EEE of each wavelength in the helium gas emission spectrum, we use the formula derived from Planck’s equation: E=hcλE = \frac{hc}{\lambda}E=λhc
Where:
- EEE = energy of the photon (in joules, J)
- hhh = Planck’s constant = 6.626×10−34 J\cdotps6.626 \times 10^{-34} \, \text{J·s}6.626×10−34J\cdotps
- ccc = speed of light = 3.00×108 m/s3.00 \times 10^8 \, \text{m/s}3.00×108m/s
- λ\lambdaλ = wavelength (in meters, so convert nm to m by multiplying by 10−910^{-9}10−9)
Helium Gas Emission Spectrum Table
| Wavelength (nm) | Wavelength (m) | Color | Energy (J) |
|---|---|---|---|
| 447 | 4.47×10−74.47 \times 10^{-7}4.47×10−7 | Violet | 4.45×10−194.45 \times 10^{-19}4.45×10−19 J |
| 471 | 4.71×10−74.71 \times 10^{-7}4.71×10−7 | Blue | 4.22×10−194.22 \times 10^{-19}4.22×10−19 J |
| 501 | 5.01×10−75.01 \times 10^{-7}5.01×10−7 | Green | 3.97×10−193.97 \times 10^{-19}3.97×10−19 J |
| 587 | 5.87×10−75.87 \times 10^{-7}5.87×10−7 | Yellow | 3.39×10−193.39 \times 10^{-19}3.39×10−19 J |
| 668 | 6.68×10−76.68 \times 10^{-7}6.68×10−7 | Red | 2.97×10−192.97 \times 10^{-19}2.97×10−19 J |
The bright-line emission spectrum of helium, like those of other noble gases, consists of distinct lines of specific colors. Each line corresponds to a photon emitted when an electron in the helium atom transitions from a higher to a lower energy level. The energy lost during this transition is emitted as electromagnetic radiation—light—with a characteristic wavelength.
To determine the energy associated with each emitted wavelength, we apply Planck’s equation. This equation relates the energy of a photon to its wavelength using two fundamental constants: Planck’s constant (hhh) and the speed of light (ccc).
Because wavelength is given in nanometers (nm), it’s crucial to convert to meters (m) to ensure consistent SI units. The shorter the wavelength, the higher the energy of the photon. This inverse relationship between energy and wavelength explains why violet light (shorter wavelength) has higher energy compared to red light (longer wavelength).
For instance, violet at 447 nm corresponds to an energy of about 4.45×10−194.45 \times 10^{-19}4.45×10−19 joules, whereas red at 668 nm corresponds to only 2.97×10−192.97 \times 10^{-19}2.97×10−19 joules. These small amounts of energy are typical for individual photons, which is why these calculations are done at such a fine scale.
In spectroscopy, identifying these emission lines is essential. They act like fingerprints for each element. Helium’s unique spectrum helps distinguish it from other gases like neon and krypton. Understanding how energy and wavelength relate allows scientists to analyze star compositions, laboratory plasmas, and even the quantum behavior of electrons in atoms.
This knowledge not only helps in scientific identification but also forms the foundation of technologies such as lasers, fluorescence analysis, and quantum optics.