# (a) Interpretation: The 100% 55 M n manganese is found naturally. The cross section of the nuclide which has electron capture is 13.3×10- 24 cm 2 . The manganese 56 M n which is a beta and gamma emitter is the product of the neutron capture. The half time of beta and gamma emitter is 2.50h.It was considered that the irradiation of sample for 1h at neutron flux 1.8×10 12 neutrons/(cm 2 s) results in the background counting rate of 10cpm. The minimum mass of manganese that could be detected is to be determined. Concept introduction: The value of minimum mass is calculated as follows: A = N ϕ σ S Or, N = A ϕ σ S Where, A = activity Ø = neutron flux σ = cross-section area S = saturation factor The saturation factor is represented as follows: S = 1 − e − 0.693 t / t 1 / 2

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

#### Solutions

Chapter 32, Problem 32.20QAP
Interpretation Introduction

## (a)Interpretation:The 100% 55Mn manganese is found naturally. The cross section of the nuclide which has electron capture is 13.3×10-24cm2. The manganese 56Mn which is a beta and gamma emitter is the product of the neutron capture. The half time of beta and gamma emitter is 2.50h.It was considered that the irradiation of sample for 1h at neutron flux 1.8×1012 neutrons/(cm2s) results in the background counting rate of 10cpm. The minimum mass of manganese that could be detected is to be determined.Concept introduction:The value of minimum mass is calculated as follows:A=NϕσSOr,N=AϕσSWhere,A = activityØ = neutron fluxσ = cross-section areaS = saturation factorThe saturation factor is represented as follows:S=1−e−0.693t/t1/2

Interpretation Introduction

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