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Measurement of energy spectra of α particles by a solid detector

Measurement of energy spectra of α particles by a solid detector. Wang Mian 6 th ,March, 2009. Energy Calibration of Si detector. E=(13.45±0.20)×Channel assuming E=0 for 0ch. Measurement of energy loss of α particle in Al. Photo of Fig.19.

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Measurement of energy spectra of α particles by a solid detector

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  1. Measurement of energy spectra of α particles by a solid detector Wang Mian 6th,March,2009

  2. Energy Calibration of Si detector E=(13.45±0.20)×Channel assuming E=0 for 0ch

  3. Measurement of energy loss of α particle in Al

  4. Photo of Fig.19 The energy loss in Fig.19 is: 0.57 MeV•mg-1•cm2 My result is: (0.56±0.02)MeV•mg-1•cm2 Fit Well

  5. Determination of Al foil thickness From Fig.19, For Eα=5.4856MeV, one can find the range R0=6.8mg/cm2 D=(R0-R’)/ρAl

  6. Question 12: How many electron hole pair will be created when α particle From 241Am enters in Si detector and losses all the energy? ω≈3ev for electron hole pair creation in Si detector N=E/ ω=5.4×106/3=1.8×106 The number of electron hole pair Whatis the expected energy solution? Expected energy solution is R=2.35×Sqrt(F/N) F: Fano Factor, is 0.72 for solid state detector R=6.0×10-4

  7. Question 13: Determine the energy resolution of the Si detector .

  8. Question 14:Plot values of full width at half maximum in each pluse-height spectrum as a function of Al sheet thickness. What is reason of this dependence?

  9. Thanks

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