Pressure, Temperature, and RMS Speed — Question 10

Section 19-3 · Physics pages 3090–3091

Question 10 Reflection and absorption produce different beam pressures A normal molecular beam has flux J = 2.00 × 1022 m−2s−1, mass m = 3.00 × 10−26 kg and speed 400 m/s. A fraction r reflects elastically and the rest sticks, without re-emission during the measurement. Measured force per area is .360 Pa. vx −vx reflect or stick Elastic reflection reverses the normal velocity component. (a) Derive pressure as a function of r and infer the reflected fraction. (b) Find the allowed pressure range and the mass accumulation rate per unit area at the inferred r. Why is this not an equilibrium gas-pressure measurement?
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Question 10 - Solution Reflection and absorption produce different beam pressures vx −vx reflect or stick (a) Method and calculation p = J[2rmv + (1 −r)mv] = Jmv(1 + r), Jmv = .240 Pa. .360 = .240(1 + r) ⇒r = .500. (b) Interpretation and further calculation 0 ≤r ≤1 ⇒.240 ≤p ≤.480 Pa. ˙mwall A = (1 −r)Jm = .500(2.00 × 1022)(3.00 × 10−26) = 3.00 × 10−4 kg/(m2s). Particles arrive in a directed beam and accumulate at the wall; there is no isotropic stationary molecular population. Physical check Absorbing a molecule transfers mv normal momentum; reversing it transfers 2mv.

Question 10

Reflection and absorption produce different beam pressures

A normal molecular beam has flux J=2.00×1022m−2s−1J=2.00\times10^{22}\,\mathrm{m^{-2}s^{-1}}, mass m=3.00×10−26m=3.00\times10^{-26} kg and speed 400 m/s. A fraction rr reflects elastically and the rest sticks, without re-emission during the measurement. Measured force per area is .360 Pa.

Diagram shown in the original SVG worksheet.

  1. Derive pressure as a function of rr and infer the reflected fraction.

  2. Find the allowed pressure range and the mass accumulation rate per unit area at the inferred rr. Why is this not an equilibrium gas-pressure measurement?

Question 10 - Solution

Reflection and absorption produce different beam pressures

Diagram shown in the original SVG worksheet.

(a) Method and calculation

p=J[2rmv+(1−r)mv]=Jmv(1+r),Jmv=.240Pa.p=J[2rmv+(1-r)mv]=Jmv(1+r),\quad Jmv=.240\,\mathrm{Pa}..360=.240(1+r)⇒r=.500..360=.240(1+r)\ \Rightarrow\ r=.500.

(b) Interpretation and further calculation

0≤r≤1⇒.240≤p≤.480Pa.0\le r\le1\ \Rightarrow\ .240\le p\le.480\,\mathrm{Pa}.ṁwallA=(1−r)Jm=.500(2.00×1022)(3.00×10−26)=3.00×10−4kg/(m2s).\frac{\dot m_{\mathrm{wall}}}A=(1-r)Jm=.500(2.00\times10^{22})(3.00\times10^{-26})=3.00\times10^{-4}\,\mathrm{kg/(m^2s)}.Particles arrive in a directed beam and accumulate at the wall; there is no isotropic stationary molecular population.

Physical check

Absorbing a molecule transfers mvmv normal momentum; reversing it transfers 2mv2mv.