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Chapter 36: Problem 32

The \(4 f \rightarrow 3 p\) transition in sodium produces a spectral line at \(567.0 \mathrm{nm} .\) Find the energy difference between these two levels.

Short Answer

Expert verified
The energy difference between the \(4 f\) and \(3 p\) levels in sodium is approximately \(3.51 \times 10^{-19} \, J\).

Step by step solution

01

Identify Given Parameters

The given wavelength for the spectral line is \(567.0 \, nm\) or \(567.0 \times 10^{-9} \, m\). The speed of light \(c\), is a known constant with a value of \(3.0 \times 10^{8} \, m/s\). Planck's constant \(h\), another known constant, which is approximately equal to \(6.63 \times 10^{-34} \, Js\).
02

Convert Wavelength to Frequency

The frequency (\(v\)) of the light can be calculated by rearranging the speed of light equation, \(c = \lambda v\), to \(v = c / \lambda\), where \(\lambda\) is the wavelength. Substituting the known values gives \(v = (3.0 \times 10^{8} \, m/s) / (567.0 \times 10^{-9} \, m) = 5.29 \times 10^{14} \, Hz\).
03

Calculate Energy Difference

The energy (\(E\)) can now be calculated using Planck's equation, \(E = hv\), substituting the calculated frequency and Planck鈥檚 constant gives \(E = (6.63 \times 10^{-34} \, Js) \times (5.29 \times 10^{14} \, Hz) = 3.51 \times 10^{-19} \, J\).

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