Intensity of radiation gah verify these answers please

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Homework Help Overview

The discussion revolves around the intensity of radiation emitted from a radioactive source, specifically examining how intensity changes with distance. The original poster presents calculations for the intensity at distances of 10 meters and 2.5 meters based on an initial intensity of 6000 Bq/m² at 5 meters.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between distance and intensity, questioning how to apply the inverse square law. There are attempts to calculate intensity at different distances, with some participants expressing confusion about the correct application of the formula.

Discussion Status

The discussion is ongoing, with participants providing guidance on the inverse relationship between distance and intensity. Some participants have attempted calculations, while others have pointed out misunderstandings regarding the constants involved in the intensity formula.

Contextual Notes

Participants are grappling with the concept of intensity as it relates to distance, with some uncertainty about how to derive the intensity at new distances based on the initial conditions provided. There is acknowledgment of the need to understand the constant in the intensity formula.

dagg3r
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A Radioactive source produces an intensity of radiation of 6000 Bq/M^2 at a distance of 5 meters from a source. Determine the intensity of the radiation at a distance of

a)10 metres

* Well i think it will be 3000 Bq/m^2 since it is further distance and is halved, can someone verify?

b) 2.5 meters

* I reckon it is 12000 Bq/M^2 since it is a closer distance from the source. someone verify?
 
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Sorry, you got them both wrong. As can be determined from the units, intensity is measured in bq/m2[/color]. The denominator should tell you that the intensity is an activity (in bq) per unit area (in m2).

Intensity thus varies inversely with the square[/color] of the distance from the source.
 
QUOTE:
Sorry, you got them both wrong. As can be determined from the units, intensity is measured in bq/m2. The denominator should tell you that the intensity is an activity (in bq) per unit area (in m2).

Intensity thus varies inversely with the square of the distance from the source.
-------------------------------------------------------------------

ok I'm not sure what you mean, but if it is the square of the distance from the source, so if the intensity is 6000bq/(m^2) and the distance is 10 metres, so does that mean the distance is 100 metres? can you please answer the first question, and i will try to answer the next or guide me through what to do because i am confused with

"Intensity thus varies inversely with the square of the distance from the source"
 
Originally posted by dagg3r
ok I'm not sure what you mean, but if it is the square of the distance from the source, so if the intensity is 6000bq/(m^2) and the distance is 10 metres, so does that mean the distance is 100 metres?

No, the distance is still 10m. What I am saying is that the intensity[/color] varies as:

I=k/r2

where k is some constant (not important for this problem). Now if you, say, double the distance, you actually decrease the intensity by a factor of 22=4.

See it now?
 
Oh I get it, ok take a look at this

so if it was 6000 bq/M^2, and the distance was 10 m,

it will be 6000/(10^2) = 6000/ 100 = 60bq/m^2
is that right?
 
Originally posted by dagg3r
so if it was 6000 bq/M^2, and the distance was 10 m,

it will be 6000/(10^2) = 6000/ 100 = 60bq/m^2
is that right?

No, because the 6000 bq/m2 is the intensity at r=5m[/color]. In your analysis, you set k=6000 bq/m2, which is incorrect.

You can do it with a ratio, if you want.

6000=k/52
x=k/102

then divide and solve. Alternatively, you could simply note that the distance doubles and do what I did in my last post.
 
ok this is what i have understood, (Your prob think I'm an idiot) well sorry


1. if i increase the distance, i decrease the intensity
2. if i decrease the distance, i increase the intensity.

ok, the first thing i learned was
6000=k/(5^2)

ok the problem i am having is you said
x=k/(10^2) ok i know that 10^2 comes from the distance squared, what i do not understand is how to obtain the intensity with I=k/r^2

right now i can't solve for x because i do not know what k is, and does 6000=k/(5^2) have anything to do with it? i know that the answer should decrease the intensity, so it should be lower than 6000 bq/m^2 so how do i do it? try to answer it but don't do the homework for me, i want to try and solve it, but i don't understand how to get I
 
wait i think i know how. check htis answer

if it is 10 metres.

i did 150000/100 = 1500 bq/m^2

if it is 2.5 metres

i did 150000/5 = 30000 bq/M^2 is that right?
 
Originally posted by dagg3r
if it is 10 metres.

i did 150000/100 = 1500 bq/m^2

Yes; when the distance doubles, the intensity decreases by a factor of 4.

if it is 2.5 metres

i did 150000/5 = 30000 bq/M^2 is that right?

No; 2.52 does not equal 5.
 

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