High-pressure sodium vapor lamps are used in street lighting. The two brightest lines in the sodium spectrum are at 589.00 and \(589.59 \mathrm{nm}\). What is the difference in energy per photon of the radiations corresponding to these two lines?

Short Answer

Expert verified
The difference in energy per photon of the radiations corresponding to these two lines can be found by first calculating the energies associated with each line using Planck's equation and then subtracting to find the difference.

Step by step solution

01

Conversion of Wavelengths to Frequencies

The first step involves converting the wavelengths of the sodium lines given in nanometers (nm) to frequencies in Hz. The speed of light (\(c\)) in a vacuum and the formula \(c = \lambda f\) can be used for this conversion, where \(\lambda\) is the wavelength and \(f\) is the frequency. Note: The speed of light \(c\) is \(3.00 × 10^8 m/s\). Since the wavelengths are in nm, they should be converted to meters: \(589.00 nm = 589.00 × 10^-9 m\), \(589.59 nm = 589.59 × 10^-9 m\). Frequencies can then be calculated as \(f_1 = c / \lambda_1\) and \(f_2 = c / \lambda_2\)
02

Calculate the Energy of Each Photon

Energies corresponding to each frequency can then be calculated using Planck’s formula \(E = hf\), where \(h\) is Planck’s constant (\(6.626 × 10^{-34} J.s\)). Therefore, the energies for the two frequencies would be \(E_1 = h f_1\) and \(E_2 = h f_2\)
03

Find the Difference in Energy

Finally, subtract the smaller energy from the larger one to get the difference in energy per photon for the two lines. \( \Delta E = E_2 - E_1 \)

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