Question

Naturally occurring manganese is $100\%{}^{55}Mn$. This nuclide has a thermal neutron capture cross section of $13.3\times {10}^{-24}c{m}^2$. The product of neutron capture is ${}^{56}Mn$, which is a beta and gamma emitter with a $2.50-h$half-life. When Figure 32-9 was produced, it was assumed that the sample was irradiated for $1h$at a neutron flux of localid="1646475257949" $1.8\times {10}^{12}neutrons/\left(c{m}^{2}s\right)$and that $10cpm$above background could be detected.

(a) Calculate the minimum mass of manganese that could be detected.

(b) Suggest why the calculated value is lower than the tabulated value.

Short answer

(a) The minimum mass of manganese that could be detected is $2.7\times {10}^{-12}g$.

(b) Because the detection coefficient of the calculated value is one and the tabulated number is in the range of $10-15\%$, the calculated value is less than the tabulated value.

Step-by-step solution

Given (part a)

Neutron capture cross section$=\sigma =13.3\times {10}^{-24}c{m}^2$

Half life$=t_{1/2}=2.5h$

Neutron flux$=\varphi =1.8\times {10}^{12}neutrons/\left(c{m}^{2}s\right)$

Time$=t=1h$

Activity$=A=10cpm$

Calculation (part a)

The saturation factor for the given example can be calculated as:

$$\begin{gathered} S=1-e^{-0.693t/t_{1/2}} \\ S=1-e^{-0.693\times 1/2.5} \\ S=0.242 \end{gathered}$$

The number of manganese atoms can be calculated as:

$A=N\varphi \sigma S$

By rearranging the terms, we get,

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

By substituting the values in the above equation, we get,

$$\begin{gathered} N=\frac{0.167}{1.8\times {10}^{12}\times 13.3\times {10}^{-24}\times 0.242} \\ N=2.9\times {10}^{10}atoms \end{gathered}$$

We know that,

Molar mass of manganese$=55g/mol$

Hence,

Mass of manganese can be calculated as:

$$\begin{gathered} Mass=\frac{No.ofatoms\times molarmass}{Avogadronumber} \\ Mass=\frac{2.9\times {10}^{10}\times 55}{6.023\times {10}^{23}} \\ Mass=2.7\times {10}^{-12}g \end{gathered}$$

Final answer (part a)

The minimum mass of manganese is calculated to be$2.7\times {10}^{-12}g$that could be detected.

Concept Introduction (part b)

Based on the data provided, the calculated value is statistically determined. The tabular value is derived from the data in the created table after it has been processed.

Explanation (part b)

The detection coefficient is the smallest difference between a statistic and a tabular number. The calculated value has a detection coefficient of one, but the tabulated value has a detection coefficient within a range of $10-15\%$. As a result, the calculated value exceeds the tabulated value.

Final answer (part b)

The calculated value is less than that of the tabulated value as the detection coefficient of the calculated value is one than that of the tabulated value which is in the range of$10-15\%$.