radiation safety test-aramco (sample q&a)2.pdf

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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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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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


= 1.5 mSv 1 Rem

hr. 10 mSv

= 0.15 Rem/hr.

Time = 2 hrs. 30 minutes

= 2.5 hrs.

We know,


= 0.15 Rem 2.5 hr.

hr.

Dose = 0.375 Rem


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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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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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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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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:


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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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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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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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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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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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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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


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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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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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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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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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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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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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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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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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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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

radiation safety test-aramco (sample q&a)2.pdf

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