X Rays and the Ordering of the Elements — Question 8

Question 8

A fluorescence yield converts a vacancy rate into a photon count

K vacancies are created at rate $R_K=2.00\times10^8\,\mathrm{s^{-1}}$. The radiative yield is $\omega_K=.40$, and .80 of radiative decays produce $K_\alpha$. A detector subtends .010 of the full solid angle and has quantum efficiency .60. Assume isotropic emission and no absorption.

1. Find the emitted $K_\alpha$ rate and detected count rate.

2. Find the expected counts in .250 s and the Poisson relative standard deviation.

Question 8 - Solution

A fluorescence yield converts a vacancy rate into a photon count

(a) Method and calculation

$$R_\alpha=R_K\omega_Kb_\alpha=(2.00\times10^8)(.40)(.80)=\boxed{6.40\times10^7\,\mathrm{s^{-1}}}.$$$$\boxed{R_d=R_\alpha(.010)(.60)=3.84\times10^5\,\mathrm{s^{-1}}}.$$

(b) Interpretation and further calculation

$$\boxed{\bar N=R_dt=9.60\times10^4}.$$$$\sigma_N=\sqrt{\bar N}\Rightarrow\boxed{\sigma_N/\bar N=1/\sqrt{96000}=.003227}.$$

Physical check

The nonradiative fraction and other radiative branches are removed before applying detector acceptance and efficiency. Poisson statistics assume independent events and negligible dead time.