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Chapter 20: Problem 12

The green color observed in aurora borealis is produced by the emission of a photon by an electronically excited oxygen atom at \(558 \mathrm{nm}\). Calculate the energy difference between the two levels involved in the emission process.

Short Answer

Expert verified
The energy difference between the two levels involved in the emission process of an oxygen atom that produces the green color observed in the Aurora Borealis is \(2.21 eV\).

Step by step solution

01

Convert Wavelength from nm to Meters

The wavelength is usually measured in nanometers (nm). But, the standard unit of wavelength to be used in the equation to calculate energy is meters (m). We know, \(1 nm = 1x10^{-9} m\) So, \(558 nm = 558 x 10^{-9} m =5.58 x 10^{-7} m.\)
02

Substitute in Planck's Equation

Planck's equation is given by, \(E=\frac{hc}{λ}\) Substitute the values of Planck's constant \(h = 6.626 x 10^{-34} Js\), speed of light \(c= 3 x 10^8 m/s\), and the calculated value of wavelength in meters from step 1, \(λ = 5.58 x 10^{-7} m\) to find the energy difference.
03

Calculate Energy Difference

Now, we can calculate the energy. It turns out to be \(E=\frac{(6.626 x 10^{-34} Js)*(3 x 10^8 m/s)}{5.58 x 10^{-7} m}\) = \(3.56 x 10^{-19} J\).
04

Convert Energy from Joules to Electron Volts

The energy is usually reported in electron volts (eV) in many cases. 1 Joule \(= 6.242 x 10^{18} eV\). So, Convert \(E\) into electron volts (eV), \(E\) = \(3.56 x 10^{-19} J * 1 J / (6.242 x 10^{18} eV)\) = \(2.21 eV\).

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