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Chapter 8: Problem 70

The yellow light from a sodium-vapor street lamp is produced by a transition of sodium atoms from a \(3 p\) state to a \(3 s\) state. If the difference in energies of those two states is \(2.10 \mathrm{eV},\) what is the wavelength of the yellow light?

Short Answer

Expert verified
The wavelength of the yellow light emitted during the transition of sodium atoms from a \(3p\) state to a \(3s\) state, with an energy difference of 2.10 eV, is \(590~\text{nm}\).

Step by step solution

01

Write down the energy-wavelength relationship

The energy-wavelength relationship can be given by the following formula: Energy = \(\dfrac{hc}{\lambda}\), where - Energy is the energy difference for the transition (2.10 eV) - h is Planck's constant \((6.626 \times 10^{-34} ~\text{Js})\) - c is the speed of light \((3.00 \times 10^8 ~\text{m/s})\) - λ is the wavelength of the light (which we are supposed to find in this problem)
02

Convert energy from eV to Joules

Given energy difference is 2.10 eV. To use this energy value in the energy-wavelength formula with SI units, we need to convert it into Joules using the conversion factor of \(1~\text{eV} = 1.602 \times 10^{-19}~\text{J}\). Energy (in Joules) = \(2.10~\text{eV} \times \dfrac{1.602 \times 10^{-19}~\text{J}}{1~\text{eV}}\) Energy (in Joules) = \(3.36 \times 10^{-19}~\text{J}\).
03

Use the energy-wavelength relationship to find the wavelength

Now, we can use the energy-wavelength formula to find the wavelength of the emitted light: \(3.36 \times 10^{-19}~\text{J}\) = \(\dfrac{(6.626 \times 10^{-34} ~\text{Js})(3.00 \times 10^8 ~\text{m/s})}{\lambda}\), Solving for λ, we get: \(\lambda = \dfrac{(6.626 \times 10^{-34} ~\text{Js})(3.00 \times 10^8 ~\text{m/s})}{3.36 \times 10^{-19}~\text{J}}\), \(\lambda = 5.90 \times 10^{-7}~\text{m}\).
04

Convert the wavelength to nanometers

To express the wavelength of the yellow light in nanometers, we will convert it from meters to nanometers using the conversion factor of \(1~\text{m} = 10^9~\text{nm}\): \(\lambda = 5.90 \times 10^{-7}~\text{m} \times \dfrac{10^9~\text{nm}}{1~\text{m}}\) \(\lambda = 590~\text{nm}\). The wavelength of the yellow light emitted during this transition of sodium atoms is 590 nm.

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