Chapter 32, Problem 32.20QAP

Interpretation Introduction

(a)

Interpretation:

The 100% ${}^{55}Mn$ manganese is found naturally. The cross section of the nuclide which has electron capture is 13.3×10-²⁴cm². The manganese ${}^{56}Mn$ 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:

$A=N\varphi \sigma S$

Or,

$N=\frac{A}{\varphi \sigma S}$

Where,

A = activity

Ø = neutron flux

σ = cross-section area

S = saturation factor

The saturation factor is represented as follows:

$S=1-e^{-0.693t/t_{1/2}}$

Interpretation Introduction

(b)

Interpretation:

The reason for the calculated value is less than tabulated value is to be determined.

Concept introduction:

The calculated value is statistically determined on the basis of the data provided. The tabulated value is obtained by the processing of data which is present in the formed table.