Intensity question (difficult)

[bonbon22]

Homework Statement

A sample of radioisotope emits 1.2 ´ 10^12 photons per second in all directions. Assuming that the area of a person is about 1 m^2, estimate the number of photons per second received by someone standing (a) 2.0 m; (b) 4.0 m; (c) 5.5 m from the source if they were accidentally exposed.

Relevant Equations

I= k/r^2

area of a sphere 2.0 m radius:

idk what to do after this point I am lost help
answer is 3.16*10^9

 

[jbriggs444]

Insert spherical cow joke here.

It seems that you have assumed that the quoted "area" of a person is their cross-sectional area in the direction of the source. But suppose instead that it refers to the total area of their skin. Suppose further that the person is spherical.

 

[haruspex]

Insert spherical cow joke here.

It seems that you have assumed that the quoted "area" of a person is their cross-sectional area in the direction of the source. But suppose instead that it refers to the total area of their skin. Suppose further that the person is spherical.

Equivalently, you can suppose the person is a tall narrow box of square horizontal section. Either way, that brings in a factor of 1/4, but to get the book answer another factor of 1/2 is needed.

 

[bonbon22]

Insert spherical cow joke here.

It seems that you have assumed that the quoted "area" of a person is their cross-sectional area in the direction of the source. But suppose instead that it refers to the total area of their skin. Suppose further that the person is spherical.

step by step if your free m8 . would help a lot ??

 

[haruspex]

step by step if your free m8 . would help a lot ??

The total area of a sphere is 4πr², but what area would it occupy in a photograph?

 

[bonbon22]

The total area of a sphere is 4πr², but what area would it occupy in a photograph?

idk what you mean by a photograph, an area of 1 m by 1m ?

 

[haruspex]

idk what you mean by a photograph, an area of 1 m by 1m ?

Viewed from the emitting source, a spherical target looks like what?

 

[bonbon22]

Viewed from the emitting source, a spherical target looks like what?

looks like a dot or it could be an area are we thinking in three dimensional plane here. is this 21 questions ? I am too stupid please explain like i just came out the womb.

 

[haruspex]

looks like a dot or it could be an area are we thinking in three dimensional plane here. is this 21 questions ? I am too stupid please explain like i just came out the womb.

When you look at the full moon in the sky what shape do you see?

 

[bonbon22]

When you look at the full moon in the sky what shape do you see?

a circle so its an area , are you trying to get your commenting up by doing this? come on bro time is of the essence

 

[haruspex]

a circle so its an area ,

So what area is being hit by radiation, as viewed from the source?

 

[bonbon22]

So what area is being hit by radiation, as viewed from the source?

1 m

 

[gneill]

What course level is this problem from? Is the problem from a textbook?

 

[haruspex]

1 m

No, as @jbriggs444 and I have pointed out, to get something approaching the book answer we must take the 1m² as being the whole surface area of the person. Radiation only falls on the side towards the source. So the shape of the person has to be considered.
We have proposed two models, a spherical person and long thin rectangular box. Both give the same answer, but conceptually I feel you may find it easier to work with the second.

Unfortunately, this still yields double the value the book gives, so I wouid say there is an error in the book.

 

[gneill]

I asked about the course level and source of the question because I'm assuming that the photons being emitted would be gamma rays (isotope emitting photons). They have an attenuation rate when traveling through a medium such as air. Although now it occurs to me that such attenuation would not be sufficient to adjust the result to the "book's" answer given the distances involved. I 'd like to know what course level the question is aimed at so we can make suitable assumptions about what factors to consider.

At close range the simple circular disk cross section approach, where the area of the circular disk is considered to be an area patch on the surface of sphere at a given distance becomes problematic due to the geometry. I thought perhaps a solid angle (steradian) approach might be more fruitful. This did not yield a result closer to the book's value. So @haruspex, I suspect that you may be correct in your surmise that the book's proffered answer is incorrect.

 

[jbriggs444]

Unfortunately, this still yields double the value the book gives,

Could it be that @bonbon22 is comparing his 2.0 meter answer with the book's 5.5 meter answer?

The book's answer reported in post #1 was 3.16 x 10^9
My answer for part c (the 5.5 meter question) rounds to 3.16 x 10^9 as well.

[I was searching for a factor of 8 and had failed to find one. But (5.5/2.0)^2 = 7.5625]

 

[bonbon22]

Could it be that @bonbon22 is comparing his 2.0 meter answer with the book's 5.5 meter answer?

The book's answer reported in post #1 was 3.16 x 10^9
My answer for part c (the 5.5 meter question) rounds to 3.16 x 10^9 as well.

[I was searching for a factor of 8 and had failed to find one. But (5.5/2.0)^2 = 7.5625]

yes this is it
i kow what i did wrong, the website i was using phrased it in a werid way so i was a bit confused thanks for all the help.
http://www.antonine-education.co.uk/Pages/Physics_5/Nuclear_Physics/NUC_04/Nuclear_4.htm
if you scroll down you can see the question

 

[jbriggs444]

My apologies for starting out down the wrong road with the digression to spherical humans.