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# Helppp aqa a2 unit 5 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!

2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

Just quoting in Amusing Elk so she can move the thread if needed
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(Original post by Amusing Elk)
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3. (Original post by Chemgawd99)
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?
Intensity relates to energy arriving per second. The question states the count rate which is simply the number of particles arriving per second. A ratio is therefore all that is needed.

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!
We calculated the ratio between the detector surface area and the sphere of emission in all directions to be:

Which means if the detector only registers 1 in 400 (1/400) of the actual gamma photons, then to get the actual rate, we need to multiply the actual count rate x the ratio of the surface areas (how many times the sphere is larger than the detector) x 400.

Which is exactly the same as the mark scheme calculation simplified.
4. the intensity is the same at all points on the sphere, so you only need the ratio of the areas.

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