Thermal de Broglie Wavelength
A stated thermal speed model must be used consistently with lambda = h/(mv). For example, using root-mean-square speed gives p_rms = square root of 3mkT and lambda_rms = h divided by that momentum.
Why this shows up in the exam
Estimating quantum behavior in gases · Comparing electron and molecule wavelengths at one temperature · Judging when wave-packet overlap may matter
Learn the idea
Typical thermal momentum grows with the square root of mT, so thermal wavelength shrinks accordingly. Heating a gas makes its particles move faster and shortens their matter waves. At the same temperature, lighter particles have longer wavelengths because their typical momentum is smaller.
🧠 Memory hook: Hotter or heavier means shorter thermal wavelength.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v_rms = sqrt(3 k T / m) — root-mean-square speed of a classical ideal-gas particle
- lambda_rms = h / sqrt(3 m k T) — de Broglie wavelength associated with rms momentum
How to approach it
- 1Convert temperature to kelvin
- 2Use the exact speed convention named in the question
- 3Form mass-temperature ratios before inserting constants
Common slip-ups that cost marks
- •Using Celsius instead of kelvin
- •Mixing rms, mean, and most-probable speeds
- •Writing wavelength proportional to one over temperature instead of one over root temperature
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
Photons of energy 5 eV illuminate a metal of work function 2 eV. Find the stopping potential.
More from Dual Nature of Matter and Radiation
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For a photon in vacuum, energy is proportional to frequency and momentum is energy divided by c. Frequency and wavelength obey c = nu lambda, so shorter-wavelength photons have larger energy and momentum.
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For monochromatic radiation, total energy is the number of photons times h nu. Power is energy per unit time, so the photon emission rate equals power divided by single-photon energy.
Radiation Pressure and Photon Momentum Transfer
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