- #1
Sachin Vaidya
- 2
- 0
Homework Statement
Consider a evacuated long metal tube of length 'l' and diameter 'd' containing a metal source at one end. The metal source is connected to an oven and emits gaseous metal atoms in all possible directions. If an atom hits the walls of the tube, it will get stuck and will not bounce back. What fraction of the emitted atoms will be emitted out of the free end of the tube? (Neglect inter atomic collisions, consider a 2D situation)https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-xaf1/v/t1.0-9/18884_10152998608889958_7600213725768080208_n.jpg?oh=8038f95eb741b39ad8ac3bb448afcb2d&oe=55BFDFAC&__gda__=1442863826_7f94b06130ac37c9fff88440e5fc45b7
Homework Equations
The Attempt at a Solution
I consider an element dx at a distance x from the top (as shown in the figure). Atoms are emitted in all directions spanning angles from 0 to π. The atoms that will make it out must lie between the triangle shown in the figure (lines joining the element dx and the end points of the tube). So the ratio of atoms that will make it out will be the ratio of the angle subtended within the triangle and π (all possible angles).
Fraction of "useful" atoms =
fr_x=\frac{1}{\pi}\left[\tan^{-1}\left({\frac{x}{l}}\right)+\tan^{-1}\left({\frac{d-x}{l}}\right)\right]
Now how do I proceed? How do I integrate this over the whole length d? I have tried something but I'm not sure if this is correct.
F(l,d)=\frac{\frac{1}{\pi}\int_0^d \left[\tan^{-1}\left({\frac{x}{l}}\right)+\tan^{-1}\left({\frac{d-x}{l}}\right)\right]dx}{\int_0^d dx}
Which on solving becomes:
F(l,d)=\frac{2}{\pi}\left[\tan^{-1}\left(\frac{d}{l}\right)-\frac{l}{2d}\log\left(\frac{d^2}{l^2}+1\right)\right]
In the limit d<<l, this becomes almost linear:
F(l,d)=\frac{2}{\pi}\frac{d}{l}
and when l\rightarrow \infty this goes to 0.
On the other hand, in the limit d>>l, this becomes like an orifice and all the atoms are emitted out. This is reflected when the limit is taken of F(d,l) and that gives 1.