Visible (green) emission spectral line for krypton (Kr) occurs at wavelength of 533.822 nm. What is the frequency (v) of this light? (Mind the units carefully. Give your answers to six significant figures.) What is the magnitude of the energy change associated with the emission of one mole of photons of light with this wavelength?
The Correct Answer and Explanation is:
Step 1: Calculate the frequency (ν)
The frequency ν\nuν of light is related to its wavelength λ\lambdaλ by the speed of light ccc:ν=cλ\nu = \frac{c}{\lambda}ν=λc
Where:
- c=3.00×108 m/sc = 3.00 \times 10^8 \ \text{m/s}c=3.00×108 m/s (speed of light),
- λ=533.822 nm=533.822×10−9 m\lambda = 533.822 \ \text{nm} = 533.822 \times 10^{-9} \ \text{m}λ=533.822 nm=533.822×10−9 m
ν=3.00×108 m/s533.822×10−9 m=5.61883×1014 Hz\nu = \frac{3.00 \times 10^8 \ \text{m/s}}{533.822 \times 10^{-9} \ \text{m}} = 5.61883 \times 10^{14} \ \text{Hz}ν=533.822×10−9 m3.00×108 m/s=5.61883×1014 Hz
Answer (frequency): 5.61883×1014 Hz\boxed{5.61883 \times 10^{14} \ \text{Hz}}5.61883×1014 Hz
Step 2: Calculate the energy of one photon
The energy EEE of a single photon is given by Planck’s equation:E=hνE = h\nuE=hν
Where:
- h=6.626×10−34 J\cdotpsh = 6.626 \times 10^{-34} \ \text{J·s}h=6.626×10−34 J\cdotps (Planck’s constant),
- ν=5.61883×1014 Hz\nu = 5.61883 \times 10^{14} \ \text{Hz}ν=5.61883×1014 Hz
E=(6.626×10−34 J\cdotps)×(5.61883×1014 Hz)=3.72381×10−19 JE = (6.626 \times 10^{-34} \ \text{J·s}) \times (5.61883 \times 10^{14} \ \text{Hz}) = 3.72381 \times 10^{-19} \ \text{J}E=(6.626×10−34 J\cdotps)×(5.61883×1014 Hz)=3.72381×10−19 J
Step 3: Calculate energy for one mole of photons
1 mole of photons contains Avogadro’s number of photons:Emol=E×NA=(3.72381×10−19 J)×(6.022×1023)E_{\text{mol}} = E \times N_A = (3.72381 \times 10^{-19} \ \text{J}) \times (6.022 \times 10^{23})Emol=E×NA=(3.72381×10−19 J)×(6.022×1023)Emol=224,250 J/mol=2.24250×105 J/molE_{\text{mol}} = 224,250 \ \text{J/mol} = \boxed{2.24250 \times 10^5 \ \text{J/mol}}Emol=224,250 J/mol=2.24250×105 J/mol
Final Answers:
- Frequency (ν): 5.61883×1014 Hz\boxed{5.61883 \times 10^{14} \ \text{Hz}}5.61883×1014 Hz
- Energy per mole: 2.24250×105 J/mol\boxed{2.24250 \times 10^5 \ \text{J/mol}}2.24250×105 J/mol
Explanation
The electromagnetic spectrum consists of a wide range of wavelengths and frequencies, with visible light occupying only a narrow portion. Each color of visible light corresponds to a specific wavelength and frequency. In this problem, we analyze the green emission line of krypton gas, which occurs at a wavelength of 533.822 nanometers.
To determine the frequency of this light, we use the relationship between speed, frequency, and wavelength of light: c=λνc = \lambda \nuc=λν. Rearranging, we find frequency ν=c/λ\nu = c / \lambdaν=c/λ. Substituting the known values—speed of light c=3.00×108 m/sc = 3.00 \times 10^8 \ \text{m/s}c=3.00×108 m/s and wavelength converted to meters—we calculate the frequency to be approximately 5.61883×1014 Hz5.61883 \times 10^{14} \ \text{Hz}5.61883×1014 Hz, which is in the range expected for visible green light.
Next, we calculate the energy of one photon of this light using Planck’s equation: E=hνE = h\nuE=hν, where hhh is Planck’s constant. This yields an energy of about 3.72381×10−19 J3.72381 \times 10^{-19} \ \text{J}3.72381×10−19 J per photon. While this is a small amount of energy on a per-photon basis, the total energy becomes significant when considering a mole of photons.
To find the total energy associated with one mole of these photons, we multiply the energy of a single photon by Avogadro’s number, 6.022×1023 mol−16.022 \times 10^{23} \ \text{mol}^{-1}6.022×1023 mol−1. This gives a total energy release of 2.24250×105 J/mol2.24250 \times 10^5 \ \text{J/mol}2.24250×105 J/mol, which represents the magnitude of the energy change due to emission of light at this wavelength.
This kind of calculation is essential in fields like atomic spectroscopy, quantum chemistry, and photophysics, where understanding the energy of photons helps explain atomic transitions and chemical behavior
