A particular type of fundamental particle decays by transforming into an electron e− and a positron e+. Suppose the decaying particle is at rest in a uniform magnetic field of magnitude 3.53 mT and the e− and e+ move away from the decay point in paths lying in a plane perpendicular to images. How long after decay do the e− and e+ collide ?

Question

irishboy123 · Answer

weird thing is this looks velocity / energy independent

they'll both start moving opposite ways round the same circle and all things being even, ie same energy and velocity, they should meet after half a rotation

from $m \omega ^2 r = qvB = q \omega r B$

we have 

$\omega = \dfrac{qB}{m} = \dfrac{2 \pi}{T}$



seems to make some sense, but i'd tag michele if it really matters

farcher · Answer

You have found the cyclotron frequency.  
In the first cyclotron it worked because the period of revolution was independent of the energy/velocity.

ganeshie8 · Answer

$$\omega = \dfrac{qB}{m} = \dfrac{2 \pi}{T} \implies T = \dfrac{2\pi m}{qB}$$

irishboy123 · Answer

you want a half period

ganeshie8 · Answer

$$T = \dfrac{2\pi*9.1*10^{-31}}{1.6*10^{-19}*3.53*10^{-3}} \approx 10nS $$ http://www.wolframalpha.com/input/?i=%5Cdfrac%7B2%5Cpi*9.1*10%5E%7B-31%7D%7D%7B1.6*10%5E%7B-19%7D*3.53*10%5E%7B-3%7D%7D+

ganeshie8 · Answer

10/2 = 5nS

ganeshie8 · Answer

matches perfectly w/ the textbook answer ! xD

ganeshie8 · Answer

Yes @Farcher  we may assume the speeds are much less than the speed of light, so that the period/frequency are independent of velocity/energy :)

ganeshie8 · Answer

Thanks @IrishBoy123