- #1
Hercuflea
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Homework Statement
A 1 meter long uniform line source of high energy photons emits 10^7 photons /sec. Calculate the particle fluence rate (flux density) at a point P 1m away from the midpoint of the source.
Homework Equations
## dN = number~of~particles ##
## da = differential~area##
##\Phi = fluence = \frac{dN}{da}##
##\phi = fluence~rate = \frac{d\Phi}{dt} = \frac{d}{dt} \frac{dN}{da} = \frac {\frac{dN}{dt}}{da}##
The Attempt at a Solution
I have tried to solve the problem two different ways, and I get 2 different answers.
Define the origin (0,0) to be the location of the point P. Define the rod source such that it's endpoints lie at ##(1,-\frac{1}{2})## and ## (1, \frac{1}{2})##.
1st method: I integrated using the pythagorean theorem and I got
$$ r = \sqrt{y^{2} + 1^{2}}$$
$$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dr $$
$$ = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{10^7}{4\pi (\sqrt{y^2+1})^2} dy\ $$
$$ = \frac{10^7}{4\pi} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac {1}{y^2+1} dy\ $$
$$ = \frac{10^7}{4 \pi} (arctan(y)) $$ from -0.5 to 0.5
$$= \frac{10^7}{4 \pi} (arctan(0.5)-arctan(-0.5)) $$
$$= 7.379 \times 10^5 \frac{photons}{m^2 s} (\frac{1m}{100cm})^2 = 73.79 \frac{photons}{cm^2 s}$$
2nd method:
Since ##rcos(\theta) = 1 ## always, I solved for ##r## in terms of ##\theta## and I got
$$ r = \frac{1}{cos\theta}$$
Also, since ##tan\theta_{max} = \frac{\frac{1}{2}}{1} ## we have ##\theta_{max} = tan^{-1} (0.5)##.
Therefore
$$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dr $$
$$= \int_{-tan^{-1}(0.5)}^{tan^{-1} (0.5)} \frac{10^7}{4\pi {\frac{1}{cos\theta}}^2} d\theta $$
$$ = \frac{10^7}{4\pi} \int_{-tan^{-1}(0.5)}^{tan^{-1} (0.5)} cos^{2}(\theta) d\theta $$
$$ = \frac{10^7}{4\pi} \frac{1}{2} (\theta + \frac{1}{2} sin(2\theta)) $$ from ##-tan^{-1}(0.5)## to ##tan^{-1}(0.5)##
$$=\frac{10^7}{8\pi} (tan^{-1}(0.5)-(-tan^{-1}(0.5)) + \frac{1}{2} (sin(2tan^{-1}(0.5))-sin(-2tan^{-1}(0.5))))$$
$$ = 6.87 \times 10^5 \frac {photons}{m^2 s}(\frac{1m}{100cm})^2 = 68.7\frac {photons}{cm^2 s} $$
Why do the two methods give different answers? I am integrating over the same geometric object...only one way I am using the angle and the other way I am using the vertical component of r. Which method is correct?
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