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Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

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492 CHAPTER 9<br />

Solution<br />

rn =<br />

σn =<br />

Therefore,<br />

510 counts<br />

5 min<br />

102<br />

5<br />

− 2400 counts<br />

60 min<br />

40<br />

+ = 4.6.<br />

60<br />

rn = 62 ± 4.6 cpm.<br />

= 102 cpm − 40 cpm = 62 cpm<br />

Standard Deviation of a Product or a Quotient<br />

In the case of a product or a quotient of two or more numbers, each with its own<br />

standard deviation,<br />

(A ± σA) × / ÷ (B ± σB)<br />

(C ± σC ) × / ÷ (D ± σD)) = Y ± σY , (9.46)<br />

the coefficient of variation of the answer is given by<br />

σY<br />

Y =<br />

<br />

σA 2 <br />

σB<br />

2 <br />

σC<br />

2 <br />

σD<br />

2<br />

+ + + . (9.47)<br />

A B C D<br />

W Example 9.11<br />

A counting standard whose transformation rate is given as 1000 ± 30 min −1 is used<br />

<strong>to</strong> determine the efficiency of a counting system. The measured count rate is 200 ±<br />

10 min −1 . What is the efficiency of the counting system and the precision of the<br />

measurement?<br />

Solution<br />

The efficiency is<br />

200 min−1<br />

ε = = 0.2, or 20% ,<br />

−1 1000 min<br />

and the standard deviation of the efficiency, from Eq. (9.47), is calculated as<br />

<br />

2 2 30 10<br />

σε = 0.2<br />

+ = 0.012 = 1.2%.<br />

1000 200

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