Radiation Safety Test Aramco Sample Q A 2 PDF

 

RADIATION SAFETY EXAMINATION Sample Problems

1. 

An Ir-192 radioactive source with activity of 185GBq is used for radiography RHM = 0.55 Rem/Ci/mt/hr. At what distance 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  = D2² I2  = D1² Where,

I1  I2  D1   D2  

= 2750 mRem/hr. = 0.75 mRem/hr. = 1 meter = Cordon of distance, D2 to be calculated.

From above equation D2  =

I1 xD1² I2 

=

2750 x 1² mtr. 0.75

= 60.55 mtr. Cordon will be placed at a distance of 60.55 mtr.

Page 1 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

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 received in micro Sv?

Ans. Here, Dose rate 1 mSv

= 0.05 mSv/hr. = 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

3. 

An Ir-192 radioactive source gives out radiation of 1.5 mSv/hr. worked 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 hr.

1 Rem 10 mSv

= 0.15 Rem/hr. Time

We know,

= 2 hrs. 30 minutes = 2.5 hrs.

Dose

= Dose Rate x Time = 0.15 0.15 Rem 2.5 hr. hr.

Dose

= 0.375 0.375 Rem

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A radiographer has

 

RADIATION SAFETY EXAMINATION Sample Problems

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  = d 1  = I2  = d2 =

32 mSv/hr. 1m ? 48m

According to Inverse Square law; I1  x d1² = I2  x d2² I2 =

I1  x d1² d2²

=

32 x 1 48 x 48

I2 =

0.0138 mSv/hr.

In Rem, dose rate = 0.0138 mSv hr. =

1 Rem 10 mSv

0.00138 Rem/hr.

Dose rate = 1.38 x 10-3 Rem/hr. In Sv, Dose rate = 0.0138 0.0138 mSv hr.

103Sv 1 mSv

Dose rate = 13.8 Sv/hr.

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[10 mSv = 1 Rem]

 

RADIATION SAFETY EXAMINATION Sample Problems

5.  An Ir-192 radioactive source of of 185 GBq is used, RHM = 0.55 Rem/Ci/Mtr./h. A radiographer is working at 15 meter from source. If radiographer worked for 3 hours, hours, what is the dose received by him? Source strength

 

I1  = d 1 = I2  = d 2 =

= 185 GBq 37 = 5 Ci [ 37 GBq = 1Ci] = 5 x 0.55 Rem/Hr./Mtr. = 2.75 Rem/Hr./Mtr.

2.75 Rem/Hr. 1m Final dose rate 15 mtr.

According to Inverse Square Law I2  x d2² = I1  x d1² I2  = I1  x d1² d 2² = 2.75 x 1² 15 x 15

Rem/hr.

= 0.0122 Rem/hr. = 0.0122 Rem hr.

103 mRem 1 Rem

= 12.2 mRem/hr.  Dose

received

Dose received

= Dose rate x Time = 12.2mRem 3 hr. hr. = 36.6 mRem

Page 5 of 22

[ Time = 3 Hrs]

 

RADIATION SAFETY EXAMINATION Sample Problems

6.  A Ir-192 radioactive source, 10m distance 6mRem/hr. dose rate observing. If we are observing 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 = = = Again,

I1  x d1² d2² 6 x 10 x 10 20 x 20 1.5 mRem/hr.

Io = 1.5 mRem/hr. I = 0.75 mRem/hr. X = ?

We know, -0.693 x X

I = IOe  Or

HVL

x X 0.75 = 6e-0.693 HVL   -0.693 x X

Or

0.75 = 1.5 x e 

Or

0.75 = Ln 1.5 +0.693

   no.

X

0.22

X e  -0.693 0.22

= + 0.693 X 0.22 = 0.22”

of HVL = X = 0.22 HVL 0.22

=

1

Page 6 of 22

Ans.

 

RADIATION SAFETY EXAMINATION Sample Problems

7.  A radioactive source gives out radiation of 0.15 mSv/h. A radiographer has worked worked for two hours 30 minutes. What is the dose received in Rem? Ans: Given that: Dose rate = 0.15 mSv/h. Time during which radiation received = 2½ hours. To find dose received in Rem: Dose = Dose Rate x Time 1 Sv = 100 Rem or, 1000mSv = 100 Rem or, 1mSv = 100/1000 Rem = 0.1 Rem Dose rate = 0.15 mSv/hr = 0.15 mSv/hr. x 0.1 Rem/mSv  Now,

Dose received = 0.015 Rem/hr x 2.5 hours = 0.0375 Rem = 3.75x10 ² Rem Dose Received = 3.75 x 10 ² Rem. -

-

Page 7 of 22

= 0.015 Rem/hr.

 

RADIATION SAFETY EXAMINATION Sample Problems

8.  An Ir-192 radioactive source of 4440 GBq is used. RHM is 0.55 Rem/Ci/Mtr/Hr. At what distance a Radiographer will receive radiation 15 mRem/h? Solution: Given that: Source Activity RHM I2 

= = = =

Ir-192 4440 GBq 0.55 Rem/Ci/Mtr./Hr. 15 mRem/Hr.

37 GBq 4440 GBq

= 1Ci = 1Ci x 4440GBq 37 GBq = 120 Ci

We know,

3

RHM = 0.55 Rem/Ci/Mtr./Hr x 0.55 x 10 mRem/Ci/Mtr./Hr. = 550 mRem/Ci/Mtr./Hr.  For 120 Ci the radiation level at 1 Mtr. will be = 120 Ci x 550 mRem/Ci/Hr. = 66000 66000 mRem/Hr. mRem/Hr.  For finding distance; Inverse Sq. Law =

Given,

D2²

or,

D2  =

I1  = D2² I2  = D1² I1  = I2  = D1  = D2  =

66000mRem/hr. 15 mRem/hr. 1 Mtr. ?

= I1 x D1² I2 

I1  x D1² Mtr. = I2 

66000x 1² 15

 Ans.: The required distance = 66.33 Mtr. Mtr.

Page 8 of 22

Mtr. = 66.33 Mtr.

 

RADIATION SAFETY EXAMINATION Sample Problems

9.  An Ir-192 Ir-192 radioactive source of 2590 2590 GBq is used. RHF is 5.9 Rem/Ci/ft/hr. At what what distance a radiographer will receive radiation 3 mRem/h? Solution: Given that: Source Activity RHM I2 

= = = =

Ir-192 2590 GBq 5.9 Rem/Hr. 3m Rem/Hr.

 Now, we know that, 37 GBq 2590 GBq

For,

= 1 Ci = 1 Ci x 2590 GBq = 70 Ci 37 GBq RHF = 5.9 Rem/Ci/ft/hr = 5.9 x 103mRem/Ci/ft/hr 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  = D2² I2  = D1²

Where,

I1  = I2  = D1  = D2  = D2²

413000mRem/hr. 3 mRem/hr. 1 ft. ?

= I1  x D1² I2 

or,

The

D2  =

I1  x D1² Mtr. = I2 

413000x 1² 3

required distance = 371.03 required 371.03 ft.

Page 9 of 22

ft.. = 371.03 ft.

 

RADIATION SAFETY EXAMINATION Sample Problems

10. An Ir-192 radioactive source of 1110 GBq is used. RHM is 0.55 Rem/Ci/mtr/hr. Rem/Ci/mtr/hr. At what distance a radiographer will receive radiation radiati on 15 mRem/h? Solution: Given that, Source is Activity RHM I2 

= = = =

Ir-192 1110 GBq 0.55 Rem/Ci/Mtr/Hr. 15 mRem/Hr.

We know that, 37 GBq = 1 Ci 1110 GBq = 1 Ci x 1110 GBq = 30 Ci 37 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  = D2² I2  = D1² Or

D2² = I1  x D1² I2 

Or

D2 =

I1 x D1² I2 

Where, I1  = I2  = D1   = D2   = D2

 

=

16500mRem/hr. 15 mRem/hr. 1 mtr.. ?

16500 x 1² Mtr. = 33.16 Mtr. 15

The required distance = 33.16 Mtr.

Page 10 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

11. A radioactive source gives out radiation radiation of 0.6 mSv/h. A radiographer has work worked ed for two hours 30 minutes. What is the dose received in Rem? Solution: Dose received = Dose Rate x Time Given,

The

Dose Rate

= 0.6 mSv / Hr. = 0.6 mSv/Hr. x 0.1 Rem/mSv = 0.06 Rem/hr.

Time

= 2 Hrs. 30 Min. = 2½ Hr.

 Dose

= 0.06 Rem/ hr. x 2½ Hr. = 0.15 Rem

dose received by the radiographer = 0.15 Rem.

Page 11 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

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 Radiation level will be; = 4.85Ci x 0.55 0.55 Rem/Ci/Mtr./hr. Rem/Ci/Mtr./hr. = 2.6675 Rem/hr. Where, 1 = II2   = D1   = D2   =

2.6675 Rem/hr.= 0.00075 Rem/hr. 0.75 mRem/hr. 1 mtr. ?

We know I1  = D2² I2  = D1²

D2  

=

I1 x D1² I2 

=

2.6675 x 1² 0.00075

= 59.63 meter Cordon of distance = 59.63 Ans.

Page 12 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

13. A radioactive source gives gives out radiation of 0.05 mSv/hr. A radiographer has worked for three hours 30 minutes that is the dose received in i n micro Sv? Solution: We know, Dose

= Dose rate x Time Dose rate = 0.05 mSv/hr. Time = 3.5 hrs.

= = = Dose =

0.05 mSv/hr. mSv/hr. x 3.5 hrs. 0.175 mSv 175 Sv [1mSv = 1000 Sv] 175 Sv

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RADIATION SAFETY EXAMINATION Sample Problems

14. An Ir-192 Radioactive Source of 3700 m Curie Curie is used. RHM is 0.55 Rem/Ci/Mtr./hr. at what distance a radiographer will received radiation 5 mRem/hr. Here, 3700 Ci 1000

= 3.7 3.7 Ci

for 3.7 Ci at 1 meter distance the radiation Level will be = 3.7 x 0.55 = 2.035 Rem/hr. I1  = D2² I2  = D1²

I1  = 5.035 Rem/hr. I2  = 5mRem = 5 mRem =0.005Rem/hr. 1000 D1  = 1 Meter D2  = ?

D2  

=

I1 x D1² I2 

=

2.035 x I² 0.005

=

407

The received distance = 20.17 meter.

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= 20.17 meter

 

RADIATION SAFETY EXAMINATION Sample Problems

15. An Ir-192 Radioactive Source of 3700 GBq is used. RHF is 5.9 5.9 Rem/Ci/ft./hr. distance a radiographer will receive Radiation 5mRem/hr.?

At what

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 l evel will be 5.9 x 100 = 590 Rem/hr.

I1  = D2² I2  = D1²

I1  = 590 mRem/hr. I2  = 5 mRem/hr. = 5 mRem/hr. 1000 = 0.005 mRem/hr. 1 = 1 ft. D D2  = ?

D2  

=

I1 x D1² I2 

=

590 x I² 0.005

= 343.5 ft.

The distance = 343.5 ft. -------------------------------Another process, The distance =

Ci x RHM x 1000 5

=

100 x 5.9 x 1000 5

Page 15 of 22

= 343.5 ft.

 

RADIATION SAFETY EXAMINATION Sample Problems

16. An Ir-192 Radioactive Radioactive source of 740 GB GBq q is used RHM is 0.55 0.55 Rem/Ci/Mtr./Hr. At what distance 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  = D2² I2  = D1²

D2  

=

I1 x D1² I2 

=

11 x I² 1

I1  = 11 Rem/hr. I2  = 1 Rem/hr. [10 mSv mSv = 1 Rem] D1 = 1 Mtr. D2 = ?

= 3.31 Mtr.

Page 16 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

17.  A radioactive source gives out radiation of 0.8 mSv. A radiographer has worked for 4 hours 20 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 = 0.3464 Rem

[4 Hrs. 20 min. = 4.33 Hrs.]

Page 17 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

18. A Radioactive source gives out Radiation of 8500 Sv/hr. A Radiographer Radiographer has worked for 3 Hour 30 minutes. What is the dose received in Rem?

Here given

 = 8500 = 8500 Sv/hr. = 8.5 mSv/hr. 1000

We know

[1 mSv = 1000Sv]

10 mSv = 1 Rem 1 mSv

= 1 Rem 10 8.5 = 8.5 = 0.85 Rem 10

Dose

= Rate x Time

Dose

= 0.85 x 3.5

[3 hours 30 minutes = 3.5 hours] = 2.975 Rem

Page 18 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

19. A radioactive source gives out Radiation Radiation of 0.5 mSv/hr. A radiographer has worked for 4 hours 30 minutes. 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

Dose

mRem = 50 x 1000 = 50000 Rem.

= Dose Rate x Time

[4 hours 30 minutes = 4.5 hours]

= 50000 x 4.5 = 225000 Rem.

Page 19 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

20. A radiation survey meter measured a dose rate of 0.032 Sv/h from a source at a distance of one meter. What is the dose rate at a distance of 40 meter in micro Sv for Co-60 Source? We know, 1 Sv = 106 Sv  0.032  0.032Sv Sv = 0.032 x 106 Sv I1  = 0.032 x 106 Sv D1 = 1 Meter D2 = 40 Meter I2 = ?

I1  = D2² I2  = D1² Or I2  = I1 x D1² D2  I2  = 0.032 x 106 x 1² 40² = 20 Sv/hr.

Page 20 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

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 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  = D2² I2  = D1²

I1  = 10175 mRem/hr. D1 = 1 Meter D2 = 25 Meter I2 = ?

Or 1 I2  = I1 x D2D   ²

I2  = 10175 x 1² 25² = 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

Page 21 of 22

 

RADIATION SAFETY EXAMINATION Sample Problems

22. A radioactive source Ir-192 35ci exposed only 35 second. How much dose will receive at 48m long? 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  = D2² I2  = D1²  

I2  = I1 x D1² D2² = 0.187 x 1 ² (48)² = 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 µRem (microRem) 1 Sv 1 Sv

= =

1,000 mSv (milliSv) 1,000,000 µSv (microSv)

1 Sv 1 mSv 1 µSv

= = =

100 Rem 100 mRem 100 Rem

1 mRem 1 Rem

Direct Conversions: = 10 µSv = 10 mSv

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