Cyclotrons and Synchrotrons — Question 1

Section 28-6 · Physics pages 4572–4573

Question 1 Cyclotron timing matches one voltage reversal to each semicircular transit A nonrelativistic positive ion has mass 3.34 × 10−27 kg and charge 1.60 × 10−19 C in B=1.20 T. It crosses the accelerating gap twice per revolution, gaining qV0 at each crossing with V0=20.0 kV. Gap transit times are negligible. dee 1 dee 2 alternating gap voltage⃗ B into page Schematic; labels, not drawing lengths, specify the data. (a) Find the revolution frequency and the required fundamental alternating-voltage frequency. (b) Find the energy gained per revolution and the average energy-gain rate of one ion under these assumptions.
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Question 1 - Solution Cyclotron timing matches one voltage reversal to each semicircular transit dee 1 dee 2 alternating gap voltage⃗ B into page (a) Method and calculation T = 2πm qB , frev = qB 2πm = 9.14903 MHz. thalf = T/2 ⇒fRF = frev for fundamental operation. (b) Interpretation and further calculation ∆Kturn = 2qV0 = 6.40 × 10−15 J = 40.0 keV. ⟨dK dt ⟩ = 2qV0frev = 5.85538 × 10−8 W. Physical check The RF period is a full orbital period; the field polarity reverses after half of that period.

Question 1

Cyclotron timing matches one voltage reversal to each semicircular transit

A nonrelativistic positive ion has mass 3.34×10−273.34\times10^{-27} kg and charge 1.60×10−191.60\times10^{-19} C in B=1.20 T. It crosses the accelerating gap twice per revolution, gaining qV0 at each crossing with V0=20.0 kV. Gap transit times are negligible.

Diagram shown in the original SVG worksheet.

  1. Find the revolution frequency and the required fundamental alternating-voltage frequency.

  2. Find the energy gained per revolution and the average energy-gain rate of one ion under these assumptions.

Question 1 - Solution

Cyclotron timing matches one voltage reversal to each semicircular transit

Diagram shown in the original SVG worksheet.

(a) Method and calculation

T=2πmqB,frev=qB2πm=9.14903MHz.T=\frac{2\pi m}{qB},\quad f_{\mathrm{rev}}=\frac{qB}{2\pi m}=9.14903\,\mathrm{MHz}.thalf=T/2⇒fRF=frevfor fundamental operation.t_{\mathrm{half}}=T/2\Rightarrow f_{\mathrm{RF}}=f_{\mathrm{rev}}\quad\text{for fundamental operation}.

(b) Interpretation and further calculation

ΔKturn=2qV0=6.40×10−15J=40.0keV.\Delta K_{\mathrm{turn}}=2qV_0=6.40\times10^{-15}\,\mathrm J=40.0\,\mathrm{keV}.⟨dKdt⟩=2qV0frev=5.85538×10−8W.\left\langle\frac{dK}{dt}\right\rangle=2qV_0f_{\mathrm{rev}}=5.85538\times10^{-8}\,\mathrm W.

Physical check

The RF period is a full orbital period; the field polarity reverses after half of that period.