Chapter 2 Section 5 :Chapter 2 Section 5 Lauren Kuperus Intro to Nuclear Medicine Group One


Problem 2 :Problem 2 Three samples are withdrawn from a standard solution using a semi-automatic pipetting device. What is the standard deviation if the samples provide the following counts? 13967cpm, 14472cpm, 14229cpm


Calculations :Calculations 13967+14472+14229/3= 14222.66667 (13967-14223)^2= -256^2 = 65536 (14472-14223)^2= 249^2 = 62001 (14229-14223)^2= 6^2 = 36 65536+62001+36= 127573 Then take the square root of 127573/3-1 = 252


Problem 3 :Problem 3 Background counts are repeatedly acquired on a well-counter. What is the standard deviation of the values obtained? 127cpm, 112cpm, 143cpm, 133cpm, 147cpm, 159cpm.


Calculations :Calculations 127+112+143+133+147+ 159/6= 137 (127-137)^2= -10^2= 100 (112-137)^2= -25^2= 625 (143-137)^2= 6^2= 36 (133-137)^2= -4^2= 16 (147-137)^2= 10^2= 100 (159-137)^2= 22^2= 484


Continued :Continued 100+ 625+ 36+ 16+ 100+ 484= 1361 Take the square root of 1361/6-1= 16


Chapter 3Section 4Problem #1 :Chapter 3Section 4Problem #1 Nancy Hux NMT 1002 Group 1


A radioactive source produces 160mrem/hr @ 1meter. What is the exposure rate at 3meters? :A radioactive source produces 160mrem/hr @ 1meter. What is the exposure rate at 3meters? To solve: (160mrem/h)(1m)2 = (3m)2 (x) X= (160mrem/h)/(3m)2 = 160/9 =


Answer :Answer 17.77 or 18 mrem/h


Chapter 3Section 4Problem 7 :Chapter 3Section 4Problem 7 Nancy Hux NMT 1002 Group 1


If a source produces an exposure rate of 50mrem/h @ 3 inches from the source, how far away must you stand to reduce the exposure rate to 0.5 mrem/h?How far away must you stand from the source in the above problem to decrease the exposure rate to 0.05mrem/h? :If a source produces an exposure rate of 50mrem/h @ 3 inches from the source, how far away must you stand to reduce the exposure rate to 0.5 mrem/h?How far away must you stand from the source in the above problem to decrease the exposure rate to 0.05mrem/h?


Slide 12:(3in)2 (50mrem/h) = (x)2 (0.05mrem/h) X’2 = (3in)2 (50mrem/h)/ 0.05mrem = 450/0.05 = 900 (squared) =


Answer :Answer 94.868 or X = 95 inches


Unit dose adjustmentJoel ThompsonGroup One :Unit dose adjustmentJoel ThompsonGroup One A calculation to figure out how much to excess to discard from a syringe.


Slide 15:A unit dose of tc99 is caliberated to contain 20mCi in 1.2 at 2pm. A 20 mCi dose is needed at 10 am. How many ml must be discarded so the correct dose remains in the syringe.


Slide 16:4 hr pre cal factor = 1.59 20mcl x 1.59 =31.72mcl 31.72mcl / 1.2ml =26.43mcl/ml 20mci needed / 26.43mcl/ml =.756ml 1.2ml - .756ml = .44ml discard


Slide 17:A unit dose of tc99 is calibrated to contain 20 mciin .5 ml at 2pm. At 12 noon a 10mci dose is needed. How many ml must be discarded from syringe for correct dose?


Slide 18:2 hr precal=1.26 20 mci x 1.26= 25.2 25.2 / .5ml= 50.4mci/ml 10mci needed 50.4 /= .198mci/ml .5ml - .198mci/ml = .302 discard


Speaker notes :Speaker notes Find precal hrs X mci at time by precal Then divide by ml giving the concentration for that time. Divide the mci that needs to be injected by the mci/ml answer we got. This is how much we need to administer. Subtract starting ml from end figure if it asks for the discard amount.


Chapter 5 Section 11 :Chapter 5 Section 11 Sean Feinberg Intro to Nuclear Medicine Group One


Objectives :Objectives How to calculate pediatric doses using Clark’s formula


Clark’s Formula :Clark’s Formula What is Clark’s formula? It’s a formula that basically allows calculation of pediatric doses by comparing the child’s weight to an average adult weight of 150 lb. The usual standard adult dose is adjusted accordingly. Clark’s formula: child’s dose= (child’s weight in kg) (adult dose)/150lb


Clark’s Formula :Clark’s Formula Section # 11 , Question #5 If the adult dose for a GI bleeding scan is 20mCi, how many mCi should be administered to a 75lb child? Cd= (75lb)(20mCi)/(150lb) Final Answer: 10mCi


Clark’s Formula :Clark’s Formula Section #11, Question #10 At 2:00p.m. a vial of Tc99m MDP contains 92.8 mCi in 4.7ml. The facility’s adult dose is 20 mCi. At 4:30p.m. a dose is needed for a 53lb child. What dose must be used? What volume must be administered?


Clark’s Formula :Clark’s Formula Section #11, Question #10 Cd= (53lb)(20mCi)/150lb= 7.061mCi Cd= 7.1 mCi DF (decay factor) = e -.693 (2.5h)/(6.01h) DF= .7495 SC (specific concentration)= 92.8mCi/4.7ml SC= (19.74mCi/ml) (.7495) = 14.8mCi/ml V (volume)= (7.1mCi)/(14.8mCi/ml) V= .5ml Final Answer: 7.1mCi is needed & .5ml is to be admin.


Quiz :Quiz How is Clark’s formula expressed? If the adult dose for a bone scan is 22mCi, what dose should be given to a 50lb child? A 140 lb child needs a DTPA renogram. How many mCi should be administered if the adult dose is 20mCi? If 300uCi of I123 are given to adult patients, how many would be given to 60lb child? Standard adult dose scan is 8mCi, how many mCi would be administered to a 20lb child?


Answers :Answers child’s dose= (child’s weight in kg) (adult dose)/150b 7.3mCi 19mCi 120umCi 1.1mCi


References :References www.dictionary.com Nuclear Medicine and Pet, fifth edition by, Christian Bernier Basic Math Skills for NMT workbook