Chapter 2: Problem 57
For reasons having to do with quantum mechanics, a given kind of atom can emit only certain wavelengths of light. These spectral lines serve as a "fingerprint." For instance, hydrogen's only visible spectral lines are \(656,486.434,\) and \(410 \mathrm{nm} .\) If spectral lines were of absolutely precise wavelength, they would be very difficult to discern. Fortunately, two factors broaden them: the uncertainty principle (discussed in Chapter 4 ) and Doppler broadening. Atoms in a gas are in motion, so some light will arrive that was emitted by atoms moving toward the observer and some from atoms moving away. Thus, the light reaching the observer will cover 8 range of wavelengths. (a) Making the assumption that atoms move no faster than their rms speed-given by \(v_{\mathrm{nns}}=\sqrt{2 k_{\mathrm{B}} T / m},\) where \(k_{\mathrm{B}}\) is the Boltanann constant obtain a formula for the range of wavelengths in terms of the wavelength \(\lambda\) of the spectral line, the atomic mass \(m,\) and the temperature \(T\). (Note: \(\left.v_{\text {rms }} \ll c .\right)\) (b) Evaluate this range for the 656 nm hydrogen spectral line, assuming a temperature of \(5 \times 10^{4} \mathrm{~K}\).
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