Can you help me do part (b)(ii), this question was asked here 17 years ago but doesn't really answer this section.

Potassium-42 decays with a half-life of 12 hours. When potassium-42 decays it emits β– particles and gamma rays. One freshly prepared source has an activity of 3.0 × 107 Bq.

To determine the dose received by a scientist working with the source the number of gamma ray photons incident on each cm2 of the body has to be known. One in every five of the decaying nuclei produces a gamma ray photon.

(a) A scientist is initially working 1.50 m from the fresh source with no shielding. Show that at this

time approximately 21 gamma ray photons per second are incident on each cm2 of the scientist's body.

(b) (i) The scientist returns 6 hours later and works at the same distance from the source.

(i) Calculate the new number of gamma ray photons incident per second on each cm2 of the scientist's body. (Answer to this is 15counts per sec per cm-2)

(ii) At what distance from the source could the scientist now work and receive the original dose of 21 photons per second per cm2

Potassium-42 decays with a half-life of 12 hours. When potassium-42 decays it emits β– particles and gamma rays. One freshly prepared source has an activity of 3.0 × 107 Bq.

To determine the dose received by a scientist working with the source the number of gamma ray photons incident on each cm2 of the body has to be known. One in every five of the decaying nuclei produces a gamma ray photon.

(a) A scientist is initially working 1.50 m from the fresh source with no shielding. Show that at this

time approximately 21 gamma ray photons per second are incident on each cm2 of the scientist's body.

(b) (i) The scientist returns 6 hours later and works at the same distance from the source.

(i) Calculate the new number of gamma ray photons incident per second on each cm2 of the scientist's body. (Answer to this is 15counts per sec per cm-2)

(ii) At what distance from the source could the scientist now work and receive the original dose of 21 photons per second per cm2

I got a distance of 1.26m. Essentially, I used the formula for the other two parts of the question:

I = intensity at source / 4πr^2 (inverse square law). The intensity that the scientist wants to experience (or count rate per m^2 etc) is the same as 21 per cm2 or 212206.59 per m2 (if ur being very specific) . The intensity at the source is also known ie. The activity in part b divided by 5 which is about 4.24x10^6. U want to find the distance which is your only unknown r in the equation. Plugging the two intensities in gives you 212206.59 = 4.24x10^6/4πr^2. Rearrange this and u end up with r = 1.26 m. A quick sense check—seems about right because ur expecting it to be closer as the source is weaker now. Its also important to keep units consistent, I changed the cm2 to m2 first. Hope this helps and please let me know if my explanation is rubbish lol

I = intensity at source / 4πr^2 (inverse square law). The intensity that the scientist wants to experience (or count rate per m^2 etc) is the same as 21 per cm2 or 212206.59 per m2 (if ur being very specific) . The intensity at the source is also known ie. The activity in part b divided by 5 which is about 4.24x10^6. U want to find the distance which is your only unknown r in the equation. Plugging the two intensities in gives you 212206.59 = 4.24x10^6/4πr^2. Rearrange this and u end up with r = 1.26 m. A quick sense check—seems about right because ur expecting it to be closer as the source is weaker now. Its also important to keep units consistent, I changed the cm2 to m2 first. Hope this helps and please let me know if my explanation is rubbish lol

wait whats that formula? I = intensity at source / 4πr^2

never seen it before?

thanks a lot. this is the correct answer too!

never seen it before?

thanks a lot. this is the correct answer too!

Original post by alaziziomar_

wait whats that formula? I = intensity at source / 4πr^2

never seen it before?

thanks a lot. this is the correct answer too!

never seen it before?

thanks a lot. this is the correct answer too!

I think its probably a version of the formula but I will have a look to see if I can find some links explaining it properly

thanks!!

This one explains it in terms of light rather than gamma rays but its kind of the same thing https://www.toppr.com/guides/physics-formulas/inverse-square-law-formula/#:~:text=What%20is%20the%20Inverse%20Square,multiplied%20by%201%2Fd2.

Original post by Ldk458

This one explains it in terms of light rather than gamma rays but its kind of the same thing https://www.toppr.com/guides/physics-formulas/inverse-square-law-formula/#:~:text=What%20is%20the%20Inverse%20Square,multiplied%20by%201%2Fd2.

thank you so much!

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