Some Properties of Atoms — Question 6

Section 40-1 · Physics pages 6582–6583

Question 6 Electron impact can excite an atom only above a recoil-adjusted threshold An electron of mass m strikes a stationary atom of mass M. The atom gains internal energy ∆. At threshold the final electron and atom share one velocity. Assume nonrelativistic kinematics and neglect the change of atomic mass due to excitation. m, v M V threshold: common velocity Schematic; labels, not drawing lengths, specify the data. (a) Derive the incident kinetic-energy threshold. (b) Evaluate its excess above ∆for M/m = 4.00 × 104 and ∆= 8.00 eV.
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Question 6 - Solution Electron impact can excite an atom only above a recoil-adjusted threshold m, v M V threshold: common velocity (a) Method and calculation pi = mv, V = mv M + m, Kf = p2 i 2(M + m) = m M + mKi. Ki = ∆+ Kf ⇒Kth = ∆(1 + m/M) . (b) Interpretation and further calculation Kth −∆= ∆m/M = 8.00/(4.00 × 104) eV. Kth −∆= 2.00 × 10−4 eV . Physical check Threshold requires the smallest final translational energy consistent with momentum conservation. A very heavy target makes the recoil correction small but not exactly zero.

Question 6

Electron impact can excite an atom only above a recoil-adjusted threshold

An electron of mass m strikes a stationary atom of mass M. The atom gains internal energy Δ\Delta. At threshold the final electron and atom share one velocity. Assume nonrelativistic kinematics and neglect the change of atomic mass due to excitation.

Diagram shown in the original SVG worksheet.

  1. Derive the incident kinetic-energy threshold.

  2. Evaluate its excess above Δ\Delta for M/m=4.00×104M/m=4.00\times10^4 and Δ=8.00eV\Delta=8.00\,\mathrm{eV}.

Question 6 - Solution

Electron impact can excite an atom only above a recoil-adjusted threshold

Diagram shown in the original SVG worksheet.

(a) Method and calculation

pi=mv,V=mvM+m,Kf=pi22(M+m)=mM+mKi.p_i=mv,\quad V=\frac{mv}{M+m},\quad K_f=\frac{p_i^2}{2(M+m)}=\frac{m}{M+m}K_i.Ki=Δ+Kf⇒Kth=Δ(1+m/M).K_i=\Delta+K_f\Rightarrow\boxed{K_{\mathrm{th}}=\Delta(1+m/M)}.

(b) Interpretation and further calculation

Kth−Δ=Δm/M=8.00/(4.00×104)eV.K_{\mathrm{th}}-\Delta=\Delta m/M=8.00/(4.00\times10^4)\,\mathrm{eV}.Kth−Δ=2.00×10−4eV.\boxed{K_{\mathrm{th}}-\Delta=2.00\times10^{-4}\,\mathrm{eV}}.

Physical check

Threshold requires the smallest final translational energy consistent with momentum conservation. A very heavy target makes the recoil correction small but not exactly zero.