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7/27/2019 radiation safety test-aramco (sample q&a)2.pdf
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 1 of 22
1. An Ir-192 radioactive source with activity of 185GBq is used for radiography RHM = 0.55Rem/Ci/mt/hr. At what distance a cordon will be placed?
Solution:
Activity = 185 GBq
RHM = 0.55 rem/Ci/mtr/hr.
We know that,
37 GBq = 1 Ci
185 GBq = 1 Ci x 185 GBq = 5 Ci
37 GBq
Now, for 5 Ci the radiation level at 1 mtr. Will be
5 Ci x 0.55 rem/Ci/hr = 2.75 rem/hr.
= 2.75 x 1000 mRem/hr. = 2750mRem/hr.
Also, we know that the radiation level at cordon of distance is 0.75 mRem/hr.
From Inverse Sq. Law:
I1 = D2I2 = D1
Where, I1 = 2750 mRem/hr.
I2 = 0.75 mRem/hr.
D1 = 1 meter
D2 = Cordon of distance, D2 to be calculated.
From above equation
D2 = I1 xD1 = 2750 x 1 mtr.
I2 0.75
= 60.55 mtr.
Cordon will be placed at a distance of 60.55 mtr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 2 of 22
2. A radioactive source gives out radiation of 0.05 mSv/h. A radiographer has worked for
three hours 30 minutes. What is the dose received in micro Sv?
Ans.
Here,
Dose rate = 0.05 mSv/hr.
1 mSv = 1000 Sv
0.05 mSv = 0.05 x 1000 Sv
Dose Rate = 50 Sv/hr.
Radiographer worked for 3.5 hrs:
Dose = Dose Rate x Time
= 50 Sv/hr. x 3.5 hr.
= 175 Sv/hr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 3 of 22
3. An Ir-192 radioactive source gives out radiation of 1.5 mSv/hr. A radiographer hasworked for two hours 30 minutes. What is dose received in Rem?
Ans.:
Note: 10 mSv = 1 Rem
Dose rate = 1.5 mSv/hr.
= 1.5 mSv 1 Rem
hr. 10 mSv
= 0.15 Rem/hr.
Time = 2 hrs. 30 minutes
= 2.5 hrs.
We know,
Dose = Dose Rate x Time
= 0.15 Rem 2.5 hr.
hr.
Dose = 0.375 Rem
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 4 of 22
4. An radiation survey meter measured a dose rate of 32 mSv/hr. from a distance of one meter.What is the dose rate at a distance of 48 meter in Rem and micro Sv?
I1 = 32 mSv/hr.
d1 = 1 m
I2 = ?
d2 = 48m
According to Inverse Square law;
I1 x d1 = I2 x d2
I2 = I1 x d1
d2
= 32 x 1
48 x 48
I2 = 0.0138 mSv/hr.
In Rem, dose rate = 0.0138 mSv 1 Remhr. 10 mSv
= 0.00138 Rem/hr.
Dose rate = 1.38 x 10-3 Rem/hr.In Sv, Dose rate = 0.0138 mSv 103Sv [10 mSv = 1 Rem]
hr. 1 mSv
Dose rate = 13.8 Sv/hr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 5 of 22
5. An Ir-192 radioactive source of 185 GBq is used, RHM = 0.55 Rem/Ci/Mtr./h. Aradiographer is working at 15 meter from source. If radiographer worked for 3 hours, what
is the dose received by him?
Source strength = 185 GBq
37
= 5 Ci [37 GBq = 1Ci]= 5 x 0.55 Rem/Hr./Mtr.
= 2.75 Rem/Hr./Mtr.
I1 = 2.75 Rem/Hr.
d1 = 1 m
I2 = Final dose rated2 = 15 mtr.
According to Inverse Square Law
I2 x d2 = I1 x d1
I2 = I1 x d1
d2
= 2.75 x 1 Rem/hr.15 x 15
= 0.0122 Rem/hr.
= 0.0122 Rem 103 mRem
hr. 1 Rem
= 12.2 mRem/hr.
Dose received = Dose rate x Time
= 12.2mRem 3 hr. [ Time = 3 Hrs]
hr.
Dose received = 36.6 mRem
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 6 of 22
6. A Ir-192 radioactive source, 10m distance 6mRem/hr. dose rate observing. If we areobserving 0.75 mRem/hr. dose rate at 20m. How many half value we have to put between
the source and specimen? HVL thickness of lead is 0.22 inch.
Here
I1 = 6 mRem/Hr.
d1 = 10 Meters
d2 = 20 meters
I2 = ?
X = ?
HVL = 0.22
According to Inverse Square Law;
I2 = I1 x d1
d2
= 6 x 10 x 10
20 x 20
= 1.5 mRem/hr.
Again, Io = 1.5 mRem/hr.
I = 0.75 mRem/hr.
X = ?
We know,
I = IOe
Or 0.75 = 6e
Or 0.75 = 1.5 x e
Or 0.75 = Ln e
1.5
+0.693 = + 0.693 X
0.22
X = 0.22
no. of HVL = X = 0.22 = 1 Ans.
HVL 0.22
-0.693 x XHVL
-0.693 X0.22
-0.693 x XHVL
-0.693 x X0.22
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7/27/2019 radiation safety test-aramco (sample q&a)2.pdf
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 7 of 22
7. A radioactive source gives out radiation of 0.15 mSv/h. A radiographer has worked for twohours 30 minutes. What is the dose received in Rem?
Ans:
Given that:
Dose rate = 0.15 mSv/h.
Time during which radiation received = 2 hours.
To find dose received in Rem:
Dose = Dose Rate x Time
1 Sv = 100 Rem
or, 1000mSv = 100 Remor, 1mSv = 100/1000 Rem = 0.1 Rem
Dose rate = 0.15 mSv/hr = 0.15 mSv/hr. x 0.1 Rem/mSv = 0.015 Rem/hr.
Now, Dose received = 0.015 Rem/hr x 2.5 hours
= 0.0375 Rem
= 3.75x10- Rem
Dose Received = 3.75 x 10- Rem.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 8 of 22
8. An Ir-192 radioactive source of 4440 GBq is used. RHM is 0.55 Rem/Ci/Mtr/Hr. At whatdistance a Radiographer will receive radiation 15 mRem/h?
Solution:
Given that:
Source = Ir-192
Activity = 4440 GBq
RHM = 0.55 Rem/Ci/Mtr./Hr.
I2 = 15 mRem/Hr.
We know,
37 GBq = 1Ci4440 GBq = 1Ci x 4440GBq
37 GBq
= 120 Ci
RHM = 0.55 Rem/Ci/Mtr./Hr x 0.55 x 103mRem/Ci/Mtr./Hr.
= 550 mRem/Ci/Mtr./Hr.
For120 Ci the radiation level at 1 Mtr. will be
= 120 Ci x 550 mRem/Ci/Hr. = 66000 mRem/Hr.
For finding distance;
Inverse Sq. Law = I1 = D2
I2 = D1
Given, I1 = 66000mRem/hr.
I2 = 15 mRem/hr.
D1 = 1 Mtr.
D2 = ?
D2 = I1 x D1
I2
or, D2 = I1 x D1 Mtr. = 66000x 1 Mtr. = 66.33 Mtr.
I2 15
Ans.: The required distance = 66.33 Mtr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 9 of 22
9. An Ir-192 radioactive source of 2590 GBq is used. RHF is 5.9 Rem/Ci/ft/hr. At whatdistance a radiographer will receive radiation 3 mRem/h?
Solution:
Given that:
Source = Ir-192
Activity = 2590 GBq
RHM = 5.9 Rem/Hr.
I2 = 3m Rem/Hr.
Now, we know that,
37 GBq = 1 Ci2590 GBq = 1 Ci x 2590 GBq = 70 Ci
37 GBq
RHF = 5.9 Rem/Ci/ft/hr = 5.9 x 103mRem/Ci/ft/hr
For, 70 Ci source the radiation level at 1ft.will be
= 70 Ci x 5.9 x 103 mRem/Ci/hr = 413000 mRem/hr.
Now, as per Inverse Square Law:
I1 = D2
I2 = D1
Where, I1 = 413000mRem/hr.
I2 = 3 mRem/hr.
D1 = 1 ft.
D2 = ?
D2 = I1 x D1
I2
or, D2 = I1 x D1 Mtr. = 413000x 1 ft.. = 371.03 ft.
I2 3
The required distance = 371.03 ft.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 10 of 22
10.An Ir-192 radioactive source of 1110 GBq is used. RHM is 0.55 Rem/Ci/mtr/hr. At whatdistance a radiographer will receive radiation 15 mRem/h?
Solution:
Given that,
Source is = Ir-192
Activity = 1110 GBq
RHM = 0.55 Rem/Ci/Mtr/Hr.
I2 = 15 mRem/Hr.
We know that, 37 GBq = 1 Ci
1110 GBq = 1 Ci x 1110 GBq = 30 Ci37 GBq
RHM = 0.55 Rem/Ci/Mtr./hr. = 0.55 x 103 mRem/Ci/Mtr./hr.
= 550 mRem/Ci/Mtr./Hr.
For, 30 Ci source the radiation level at 1 mtr. distance will be 30 Ci x 550 mRem/Ci/hr.
= 16500 mRem/hr.
From Inverse Sq. Law
I1 = D2
I2 = D1
Or D2 = I1 x D1
I2
Or D2 = I1 x D1
I2
Where,I1 = 16500mRem/hr.
I2 = 15 mRem/hr.
D1 = 1 mtr..
D2 = ?
D2 = 16500 x 1 Mtr. = 33.16 Mtr.
15
The required distance = 33.16 Mtr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 11 of 22
11.A radioactive source gives out radiation of 0.6 mSv/h. A radiographer has worked for twohours 30 minutes. What is the dose received in Rem?
Solution:
Dose received = Dose Rate x Time
Given,
Dose Rate = 0.6 mSv / Hr. = 0.6 mSv/Hr. x 0.1 Rem/mSv
= 0.06 Rem/hr.
Time = 2 Hrs. 30 Min. = 2 Hr.
Dose = 0.06 Rem/ hr. x 2 Hr.
= 0.15 Rem
The dose received by the radiographer = 0.15 Rem.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 12 of 22
12.An Ir-192 Radioactive Source with activity of 4850 m Curie is used for Radiography, RHM= 0.55 Rem/Ci/Mtr./Hr. at what distance a cordon will be placed?
Solution:
Activity = 4850 mCi
= 4850/1000 Ci
= 4.85 Ci
RHM = 0.55 Rem/Ci/Mtr./hr.
For 4.85 Ci at 1 meter distance the Radiation level will be;
= 4.85Ci x 0.55 Rem/Ci/Mtr./hr.
= 2.6675 Rem/hr.
Where,
I1 = 2.6675 Rem/hr.
I2 = 0.75 mRem/hr. = 0.00075 Rem/hr.
D1 = 1 mtr.
D2 = ?
We knowI1 = D2
I2 = D1
D2 = I1 x D1
I2
= 2.6675 x 1
0.00075
= 59.63 meter
Cordon of distance = 59.63 Ans.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 13 of 22
13.A radioactive source gives out radiation of 0.05 mSv/hr. A radiographer has worked forthree hours 30 minutes that is the dose received in micro Sv?
Solution:
We know,
Dose = Dose rate x Time
Dose rate = 0.05 mSv/hr.
Time = 3.5 hrs.
= 0.05 mSv/hr. x 3.5 hrs.
= 0.175 mSv
= 175 Sv [1mSv = 1000 Sv]Dose = 175 Sv
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 14 of 22
14.An Ir-192 Radioactive Source of 3700 m Curie is used. RHM is 0.55 Rem/Ci/Mtr./hr. atwhat distance a radiographer will received radiation 5 mRem/hr.
Here,
3700 Ci = 3.7 Ci
1000
for 3.7 Ci at 1 meter distance the radiation
Level will be = 3.7 x 0.55
= 2.035 Rem/hr.
I1 = D2
I2 = D1 I1 = 5.035 Rem/hr.
I2 = 5mRem = 5 mRem =0.005Rem/hr.
1000
D1 = 1 Meter
D2 = ?
D2 = I1 x D1
I2
= 2.035 x I = 407 = 20.17 meter
0.005
The received distance = 20.17 meter.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 15 of 22
15.An Ir-192 Radioactive Source of 3700 GBq is used. RHF is 5.9 Rem/Ci/ft./hr. At whatdistance a radiographer will receive Radiation 5mRem/hr.?
37 GBq = 1 Ci
1 GBq = 1
37
3700 GBq = 3700 = 100 Ci
37
RHF = 5.9 Rem/Ci/ft./hr.
From 100 Ci at 1 feet distance the radiation level will be 5.9 x 100 = 590 Rem/hr.
I1 = D2 I1 = 590 mRem/hr.
I2 = D1 I2 = 5 mRem/hr. = 5 mRem/hr.
1000
= 0.005 mRem/hr.
D1 = 1 ft.
D2 = ?
D2 = I1 x D1
I2
= 590 x I = 343.5 ft.
0.005
The distance = 343.5 ft.
--------------------------------
Another process,
The distance = Ci x RHM x 1000 = 100 x 5.9 x 1000 = 343.5 ft.
5 5
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 16 of 22
16.An Ir-192 Radioactive source of 740 GBq is used RHM is 0.55 Rem/Ci/Mtr./Hr. At whatdistance a radiographer will received radiation 10 mSv?
37 GBq = 1 Ci
740 GBq = 740 Ci = 20 Ci
37
RHM = 0.55 Rem/Ci/Mtr./Hr.
For 20 Ci at 1 meter distance the radiation level will be 0.55 x 20 = 11 Rem/hr.
I1 = D2 I1 = 11 Rem/hr.
I2 = D1 I2 = 1 Rem/hr. [10 mSv = 1 Rem]
D1 = 1 Mtr.
D2 = ?
D2 = I1 x D1
I2
= 11 x I = 3.31 Mtr.
1
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 17 of 22
17. A radioactive source gives out radiation of 0.8 mSv. A radiographer has worked for 4 hours20 minutes. What is the dose received in Rem?
We know, 10 mSv = 1 Rem
0.8 mSv = 0.8 = 0.08 Rem
10
Dose = Dose rate x Time
= 0.08 x 4.33 [4 Hrs. 20 min. = 4.33 Hrs.]= 0.3464 Rem
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 18 of 22
18.A Radioactive source gives out Radiation of 8500Sv/hr. A Radiographer has worked for 3Hour 30 minutes. What is the dose received in Rem?
Here given = 8500Sv/hr.
= 8500 = 8.5 mSv/hr. [1 mSv = 1000Sv]
1000
We know 10 mSv = 1 Rem
1 mSv = 1 Rem
10
8.5 = 8.5 = 0.85 Rem
10
Dose = Rate x Time [3 hours 30 minutes = 3.5 hours]
Dose = 0.85 x 3.5 = 2.975 Rem
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 19 of 22
19.A radioactive source gives out Radiation of 0.5 mSv/hr. A radiographer has worked for 4hours 30 minutes. What is the dose received in micro Rem?
We know,
1 mSv = 100 mRem
0.5 mSv = 100 x 0.5 mRem = 50 mRem
1 mRem = 1000 Rem
50 mRem = 50 x 1000 = 50000 Rem.
Dose = Dose Rate x Time [4 hours 30 minutes = 4.5 hours]
= 50000 x 4.5
= 225000 Rem.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 20 of 22
20.A radiation survey meter measured a dose rate of 0.032 Sv/h from a source at a distance ofone meter. What is the dose rate at a distance of 40 meter in micro Sv for Co-60 Source?
We know,
1 Sv = 106Sv
0.032Sv = 0.032 x 106Sv
I1 = D2 I1 = 0.032 x 106Sv
I2 = D1 D1 = 1 Meter
D2 = 40 Meter
Or I2 = ?
I2 = I1 x D1D2
I2 = 0.032 x 106 x 1
40
= 20 Sv/hr.
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 21 of 22
21.An Ir-192 radioactive source of 18,500 milli Curie is used. RHM = 0.55 Rem/Ci/Mtr./hr.Radiographer is working at 25 meter from source. If radiographer during his duty of 8 hours
decides to work 4 hours, what is the dose received by him in mSv?
Ans.
For 18,500 milli Curie at 1 meter distance the radiation level will be:
= 18,500 x 0.55 x 1000
1000
= 10175 mRem/Hr.
As per Inverse Square Law,
I1 = D2 I1 = 10175 mRem/hr.
I2 = D1 D1 = 1 Meter
D2 = 25 Meter
Or I2 = ?
I2 = I1 x D1
D2
I2 = 10175 x 1
25
= 16.28 mRem/hr.Dose = Dose Rate x Time
= 16.28 x 4 = 65.12 mRem
So, 1 mRem = 1 mSv
100
65.120 mRem = 65.12 mSv
100
= 0.6512 mSv
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RADIATION SAFETY EXAMINATION
Sample Problems
Page 22 of 22
22.A radioactive source Ir-192 35ci exposed only 35 second. How much dose will receive at48m long? RHM of Ir-192 = 0.55 Rem/Ci/mtr/hr.
Here,
Source Activity = 35 Ci
RHM of Ir-192 = 0.55 Rem/Ci/Mtr./hr.
Dose Rate = 35 Ci x 0.55 Rem/Ci/Mtr./hr.= 19.25 Rem/hr. @ 1 mtr.
If the Source exposed 35 Sec.
Dose = 19.25 Rem x 1 hr. x 35 Min.
hr. 60 min. 60 Sec.
= 0.187 Rem @ 1 meter. I1 = 0.187 Rem/hr.
D1 = 1 Meter
D2 = 48 Meter
I2 = ?
As per Inverse Square Law,
I1 = D2
I2 = D1
I2 = I1 x D1
D2= 0.187 x 1
(48)
= 8.12295 x 10-5
Since, 1 Rem = 1000 mRem
I2 = 8.12295 x 10-5 Rem x 1000 mRem
Rem
= 0.08123 mRem Ans.
1 Rem = 1,000 mRem (milliRem)
1 Rem = 1,000,000 Rem (microRem)
1 Sv = 1,000 mSv (milliSv)
1 Sv = 1,000,000 Sv (microSv)
1 Sv = 100 Rem
1 mSv = 100 mRem
1 Sv = 100 Rem
Direct Conversions:1 mRem = 10 Sv
1 Rem = 10 mSv