Ask your homework questions to teachers and professors, meet other students, and be entered to win \$600 or an Xbox Series X 🎉Join our Discord! Numerade Educator

# A particular radioactive source produces 100. mrad of 2 - MeV gamma rays per hour at a distance of 1.0 m. (a) How long could a person stand at this distance before accumulating an intolerable dose of 1.0 rem? (b) Assuming the gamma radiation is emitted uniformly in all directions, at what distancewould a person receive a dose of 10. mrad/h from this source?

## a. 10 hb. 3.2 m

### Discussion

You must be signed in to discuss.
##### Top Physics 103 Educators ##### Andy C.

University of Michigan - Ann Arbor

LB  ### Video Transcript

okay from number 50. We have a radioactive source that's putting out 100 Millie rads. Pencil. Ah, 100 milli red Red's up per hour of to Meghan Luke Trimble. Gamma rays is gamma, and you're one meter away. And port A wants to know how long we have to stand there one meter away until you had a dose of one ram. We're given this in rad, but want this in Ram? Remember the how you can convert from one to the other. The dose in rim is the dose. In red times, the R B. That was the relative biological effect. Well, for gamma, this is just one. So it turns out your dose and ram is going to be the same as your dose and read for gamma rays. So we're This makes us simple so that we're just finding how long until you get a whole Graham. If you have 100 Millie rooms. Millie Reds. Sorry for each hour. Well, Millie is 10 later, thirds, and we move this three places. That's 30.1 red per hour. So how long would it take to get a whole red? You know, until you need 10 times that you need 10 hours to get a whole one where I can think of that as, um I guess just one divided by what one? Ellen, Part B. What distance would you need to be away from that source? Thio only get 10 Miller reds per hour rather than plug. You know, numbers in the equation here. This is just a proportional reasoning thing. Think of the relationship for a a source giving off radiation, the intensity of it at different distances would depend on this formula. And the only thing that matters that this is this, you know, intensity, how much I'm putting out and the distance away. So, basically, for this it was 100 and I wanted to only be 10 so I'm gonna have 1/10 of this. What distance do I need for this to be 1/10? This is the rest of his All stay in constant before the by the original density. Um, so manipulated equations. I need this equation divided by 10. So I need to 10 on the bottom here after I square it. So what? After a square, it will be the 10. That's going to be the square root of 10. So see how these red numbers, I added, will cancel out on that equation. So with the square root of 10 times what it originally waas, it's a skirt. 10 is 3.16 and this 3.16 of what the original or was our original. Our distance was one meter, so 3.16 times a one meter. So 3.16 meters. So I just used the idea that this was an inverse square law. If I want this to be 1/10 I needed this to be square root of 10. University of Virginia
##### Top Physics 103 Educators ##### Andy C.

University of Michigan - Ann Arbor

LB  