A Summing Up — Question 8

Section 44-14 · Physics pages 7566–7567

Question 8 Recoil kinematics makes detector sensitivity depend on the target mass For elastic nonrelativistic scattering of a dark particle of mass m on a stationary nucleus of mass M, use maximum recoil energy ER = 2µ2v2/M, with µ = mM/(m + M). Take mc2 = 50 GeV, Mc2 = 100 GeV and v/c=10−3. m v M recoil Schematic; labels, not drawing lengths, specify the data. (a) Calculate reduced mass and maximum recoil energy in keV. (b) At fixed m and v, show which target mass maximizes the maximum recoil energy.
Show solutionHide solution
Question 8 - Solution Recoil kinematics makes detector sensitivity depend on the target mass m v M recoil (a) Method and calculation µc2 = 50(100)/150 = 33.3333 GeV . ER,max = 2(33.3333)2(10−3)2/100 = 22.2222 keV . (b) Interpretation and further calculation ER = 2m2v2 M (m + M)2 , d dM M (m + M)2 = m −M (m + M)3 . M = m maximizes ER, ER,max = 1 2mv2 . Physical check At matched masses a head-on collision can transfer all projectile kinetic energy. Total detection rates also depend on nuclear coherence and thresholds.

Question 8

Recoil kinematics makes detector sensitivity depend on the target mass

For elastic nonrelativistic scattering of a dark particle of mass m on a stationary nucleus of mass M, use maximum recoil energy ER=2μ2v2/ME_R=2\mu^2v^2/M, with μ=mM/(m+M)\mu=mM/(m+M). Take mc2=50mc^2=50 GeV, Mc2=100Mc^2=100 GeV and v/c=10−310^{-3}.

Diagram shown in the original SVG worksheet.

  1. Calculate reduced mass and maximum recoil energy in keV.

  2. At fixed m and v, show which target mass maximizes the maximum recoil energy.

Question 8 - Solution

Recoil kinematics makes detector sensitivity depend on the target mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

μc2=50(100)/150=33.3333GeV.\boxed{\mu c^2=50(100)/150=33.3333\ {\mathrm{GeV}}}.ER,max=2(33.3333)2(10−3)2/100=22.2222keV.\boxed{E_{R,\max}=2(33.3333)^2(10^{-3})^2/100=22.2222\ {\mathrm{keV}}}.

(b) Interpretation and further calculation

ER=2m2v2M(m+M)2,ddMM(m+M)2=m−M(m+M)3.E_R=2m^2v^2\frac{M}{(m+M)^2},\quad\frac{d}{dM}\frac{M}{(m+M)^2}=\frac{m-M}{(m+M)^3}.M=mmaximizes ER,ER,max=12mv2.\boxed{M=m\ \text{maximizes }E_R,\quad E_{R,\max}=\tfrac12mv^2}.

Physical check

At matched masses a head-on collision can transfer all projectile kinetic energy. Total detection rates also depend on nuclear coherence and thresholds.