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AQA Unit 5 - Radioactivity question Watch

1. Hello!

I am currently doing the Unit 5 paper from June 2012 and have run into some trouble with question 3 of section A (nuclear and thermal physics). I have posted the following parts below along with my queries about them! I'm sure they are all pretty straightforward, but any help would be greatly appreciated!

3 (b) A γ ray detector with a cross-sectional area of 1.5 × 10–3 m^2 when facing the source is placed 0.18 m from the source. A corrected count rate of 0.62 counts s^–1 is recorded.

3 (b) (i) Assume the source emits γ rays uniformly in all directions. Show that the ratio number of γ photons incident on detector/number of γ photons produced by source is about 4 × 10–3.
The mark scheme states that the ratio of the area of the detector to the surface area of the sphere (created by emission of gamma rays, because they are given off in all directions, with radius 0.18m) is the calculation that needs to be done. I can understand this at some level, but how come intensity isn't taken into account here?

3 (b) (ii) The γ ray detector detects 1 in 400 of the γ photons incident on the facing surface of the detector. Calculate the activity of the source. State an appropriate unit.
The mark scheme says:
activity = 0.62(count rate)/(0.00368 x 1/400) = 67000Bq, where 0.00368 is a more precise value for the ratio calculated in (b) (i). I can't seem to deduce where this equation is coming from!

Thank you in advance!
2. Posting again in an attempt to 'bump' the thread up!
3. (Original post by elf03)
Hello!

I am currently doing the Unit 5 paper from June 2012 and have run into some trouble with question 3 of section A (nuclear and thermal physics). I have posted the following parts below along with my queries about them! I'm sure they are all pretty straightforward, but any help would be greatly appreciated!

3 (b) A γ ray detector with a cross-sectional area of 1.5 × 10–3 m^2 when facing the source is placed 0.18 m from the source. A corrected count rate of 0.62 counts s^–1 is recorded.

3 (b) (i) Assume the source emits γ rays uniformly in all directions. Show that the ratio number of γ photons incident on detector/number of γ photons produced by source is about 4 × 10–3.
The mark scheme states that the ratio of the area of the detector to the surface area of the sphere (created by emission of gamma rays, because they are given off in all directions, with radius 0.18m) is the calculation that needs to be done. I can understand this at some level, but how come intensity isn't taken into account here?

3 (b) (ii) The γ ray detector detects 1 in 400 of the γ photons incident on the facing surface of the detector. Calculate the activity of the source. State an appropriate unit.
The mark scheme says:
activity = 0.62(count rate)/(0.00368 x 1/400) = 67000Bq, where 0.00368 is a more precise value for the ratio calculated in (b) (i). I can't seem to deduce where this equation is coming from!

Thank you in advance!
putting a factor of 1/400 on the bottom of the fraction is equivalent to multiplying the count rate by 400.

the detector only detects 1/400 photons so multiplying the count rate by 400 gives you the number of photons going into the detector.per second
4. (Original post by Joinedup)
putting a factor of 1/400 on the bottom of the fraction is equivalent to multiplying the count rate by 400.

the detector only detects 1/400 photons so multiplying the count rate by 400 gives you the number of photons going into the detector.per second
I see! Thank you!
5. Hi, don't know if you looked at this question anymore??

Do you by any chance now why you divide it by 0.00368?

Good luck for tomorrow!
6. (Original post by Danikaschrader)
Hi, don't know if you looked at this question anymore??

Do you by any chance now why you divide it by 0.00368?

Good luck for tomorrow!
Sorry for not getting back to you in time! I hope the exam went well for you!

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