Research Article

Two New Reference Materials Based on Tobacco Leaves: Certification for over a Dozen of Toxic and Essential Elements

Table 1

Homogeneity testing of the candidate reference materials: INCT-OBTL-5 (a) and INCT-PVTL-6 (b) for 100 mg sample by NAA using Fisher’s test and Student’s t-test.
(a)

Element
(γ line, keV)
𝑥 1 ± 𝑠 1
mg kg−1
n 𝑥 2 ± 𝑠 2
mg kg−1
m F F 0.05t t 0.05

Ce (145.4) 3 . 0 9 ± 0 . 2 1 6 3 . 0 8 ± 0 . 3 6 6 2 . 0 9 8 < 5 . 0 5 0 2 . 0 6 8 < 2 . 2 2 8
Co (1332.5) 0 . 9 6 2 ± 0 . 0 2 8 6 0 . 9 9 1 ± 0 . 0 1 9 6 2 . 0 9 8 < 5 . 0 5 0 2 . 0 6 8 < 2 . 2 2 8
Cs (795.9) 0 . 2 3 9 ± 0 . 0 1 6 6 0 . 2 4 0 ± 0 . 0 1 3 6 1 . 6 2 6 < 5 . 0 5 0 0 . 1 8 6 < 2 . 2 2 8
Eu (344.3) 0 . 0 5 7 ± 0 . 0 0 4 6 0 . 0 5 8 ± 0 . 0 0 2 6 3 . 9 5 9 < 5 . 0 5 0 0 . 5 2 7 < 2 . 2 2 8
Fe (1291.6) 1 5 3 4 ± 3 9 6 1 5 7 2 ± 2 7 6 2 . 0 5 3 < 5 . 0 5 0 1 . 9 1 1 < 2 . 2 2 8
Hf (482.2) 0 . 2 5 8 ± 0 . 0 1 7 6 0 . 2 5 9 ± 0 . 0 1 1 6 2 . 3 9 3 < 5 . 0 5 0 0 . 1 8 2 < 2 . 2 2 8
Rb (1076.6) 1 7 . 2 ± 1 . 0 6 1 8 . 0 ± 0 . 5 6 4 . 8 5 3 < 5 . 0 5 0 1 . 7 2 8 < 2 . 2 2 8
Tb (879.4) 0 . 0 4 1 ± 0 . 0 1 1 6 0 . 0 3 2 ± 0 . 0 0 8 6 1 . 7 6 2 < 5 . 0 5 0 1 . 6 4 5 < 2 . 2 2 8

(b)

Element
(γ line, keV)
𝑥 1 ± 𝑠 1
mg kg−1
n 𝑥 2 ± 𝑠 2
mg kg−1
m F F 0.05 t t 0.05

Ba (216.1) 3 7 . 1 ± 2 . 5 6 3 6 . 4 ± 2 . 9 6 1 . 3 2 1 < 5 . 0 5 0 0 . 4 0 8 < 2 . 2 2 8
Co (1332.5) 0 . 1 4 4 ± 0 . 0 0 5 6 0 . 1 4 5 ± 0 . 0 0 5 6 1 . 1 6 5 < 5 . 0 5 0 0 . 1 0 3 < 2 . 2 2 8
Cs (795.9) 0 . 0 2 3 ± 0 . 0 0 2 6 0 . 0 2 3 ± 0 . 0 0 3 6 1 . 4 7 2 < 5 . 0 5 0 0 . 1 8 6 < 2 . 2 2 8
Eu (344.3) 0 . 0 1 3 ± 0 . 0 0 2 6 0 . 0 1 3 ± 0 . 0 0 3 6 1 . 6 7 1 < 5 . 0 5 0 0 . 0 1 7 < 2 . 2 2 8
Fe (1291.6) 2 5 1 ± 8 6 2 5 7 ± 1 3 6 2 . 6 6 8 < 5 . 0 5 0 1 . 1 6 5 < 2 . 2 2 8
Hf (482.2) 0 . 1 3 9 ± 0 . 0 2 2 6 0 . 1 4 5 ± 0 . 0 2 1 6 1 . 1 2 9 < 5 . 0 5 0 0 . 4 8 0 < 2 . 2 2 8
Rb (1076.6) 5 . 7 8 ± 0 . 5 0 6 5 . 8 3 ± 0 . 3 2 6 2 . 5 0 9 < 5 . 0 5 0 0 . 2 2 1 < 2 . 2 2 8
Sc (889.3) 0 . 0 5 8 ± 0 . 0 0 4 6 0 . 0 5 9 ± 0 . 0 0 2 6 3 . 3 1 3 < 5 . 0 5 0 0 . 3 0 6 < 2 . 2 2 8

𝐹 = 𝑠 2 1 ( 2 ) / 𝑠 2 2 ( 1 ) , F 0.05—critical value of the Fisher’s test at significance level α = 0.05 and degrees of freedom 𝑓 1 = 𝑓 2 = 5.
𝑡 = [ | 𝑥 1 𝑥 2 | / ( 𝑛 1 ) 𝑠 2 1 + ( 𝑚 1 ) 𝑠 2 2 ] 𝑛 𝑚 ( 𝑛 + 𝑚 2 ) / ( 𝑛 + 𝑚 ) , 𝑡 0 . 0 5 —critical value of the Student’s t-test at significance level α = 0.05 and degrees of freedom f = n + m−2 = 10.