1953
DOI: 10.1103/physrev.89.715
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The Measurement and Interpretation of theKAuger Intensities ofSn113,

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Cited by 143 publications
(22 citation statements)
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“…The efficiencies were determined by counting aliquots of the supernatant solution during and after deposition. Although this method of measurement undoubtedly gives the fraction deposited in the course of electrolysis, it does not necessarily give the fraction remaining on the fdm because dissolution of a metallic deposit after electrodeposition or during washing could lead to significant losses for some metals, as we found and as has been reported by Broyles et al (3) in the case of tin. This point is discussed by Lingane (4) in his book on electroanalytical chemistry, which is an excellent source of information on practical problems associated with electrolysis.…”
Section: Introductionsupporting
confidence: 75%
See 1 more Smart Citation
“…The efficiencies were determined by counting aliquots of the supernatant solution during and after deposition. Although this method of measurement undoubtedly gives the fraction deposited in the course of electrolysis, it does not necessarily give the fraction remaining on the fdm because dissolution of a metallic deposit after electrodeposition or during washing could lead to significant losses for some metals, as we found and as has been reported by Broyles et al (3) in the case of tin. This point is discussed by Lingane (4) in his book on electroanalytical chemistry, which is an excellent source of information on practical problems associated with electrolysis.…”
Section: Introductionsupporting
confidence: 75%
“…Thus, other reactions must take place at the electrodes; in fact, evolution of gas is observed at the cathode and gas bubbles come out of the anolyte when shaken. This can be interpreted by the following electrode reactions taking place in acidic and basic solutions, namely the evolution of hydrogen at the cathode [I] 2H+ + 2e + Hz (acid medium) [2] 2Hz0 + 2e + H z + 20H-(basic medium) and the evolution of oxygen at the anode [3] 2Hz0 + OZ + 4H+ + 4e (acid medium) [4] 40H-+ O2 + 2 H z 0 + 4e (basic medium).…”
Section: Electrode Reactionsmentioning
confidence: 99%
“…I t was assumed, for purposes of calculation, that 1 K X-ray corresponded to 1 disintegration of Hglgi9. The K X-ray fluorescence yield was taken t o be 0.945 (10).…”
Section: Resultsmentioning
confidence: 99%
“…The X-ray peak was nicely resolvable into three components-a long-lived tail, 40 hr, and 9.5 hr, and this peak was used to measure the counting rates of the metastable and ground states. Disintegration rates were calculated assuming a fluorescence yield of 0.945 (10) and that each disintegration yielded one X-ray. The cross-section values are given in Table V and are shown in Fig.…”
Section: Resultsmentioning
confidence: 99%
“…The binding energy is released in the form of X-rays or Auger electrons, (e.g. In iodine a characteristic X-ray is emitted for about 88% of the absorptions [40]) The characteristic X-rays are re-absorbed through less tightly bound electron shells of the absorber atom. Thus, the overall effect of photoelectric absorption is the liberation of a photoelectron, which carries off most of the gamma ray energy, together with one or more low-energy electrons corresponding to absorption of the original binding energy of the photoelectron.…”
Section: P Hotoelectric Absorptionmentioning
confidence: 99%