View Full Version : Radiation, range of alpha particles
Evil dead
Feb 3, 2007, 05:28 AM
Alpha particles lose 5 x 10(-18) J of kinetic energy in each collision they make with an air molecule. An alpha particle traveling through air at STP undergoes around 5(5) ionising collisions with the air molecules for each centimeter of air they travel through. Calculate the range of the alpha particle if it starts with 4.8 x 10(-13) J?
(?) = power.
I tried every possible calculation but I cnt get a good answer.
Need help please
Nosnosna
Feb 3, 2007, 06:25 AM
You are looking for the point at which the energy of the particle is reduced to 0. So, you set up a simple linear equation:
E = (initial energy) - (energy loss per collision) * (number of collisions)
Since you're interested in the point at which the particle stops, you are looking for the number of collisions where E = 0. Once you have that number, simply convert your collisions into distance using the factor given.
Evil dead
Feb 3, 2007, 06:31 AM
I don't get it...
I got -2.44 x 10(-14).
And the answer in the book says 0.96 cm...
when you say E = 0 what does that mean? And overall how would you work out the whole thing?
Nosnosna
Feb 3, 2007, 06:55 AM
E is the energy of the particle after a number of collisions. When E = 0, it will no longer exist. I should have written it as E(n) for clarity.
So,
E(n) = (initial energy) - (energy loss per collision) * n, where n is the number of collisions.
Since you're looking for the endpoint, you have
0 = 4.8 * 10(-13) - 5 * 10(-18) * n
Also written as:
n = (4.8 * 10(-13)) / (5 * 10(-18))
Once you calculate n (I'll give you a hint: you're looking at 10(4) for the exponent part... if that doesn't make sense, we'll need to go back to operations on exponents :)) you need to convert that into a distance... you're given a number of collisions per centimeter, and n is the number of collisions.
Evil dead
Feb 3, 2007, 07:29 AM
Ahhh. Thanks a bunch. I got 960, but since it goes through 10(4) collisions (10(5) originally but we have cancelled out 1 exponent) I have to divide that amount by 1000:)
Thanks a bunch.
1 more request. All this of making equations and specific rules to solve a problem, how do you do it? I want to start thinking like you do. Any tips?
The way you came up with that constant I thought is amazing:D
Nosnosna
Feb 4, 2007, 10:25 AM
It's not specific rules for the problem. It's just applying existing rules in a way that fits the problem at hand. And the equation in this case is simply y = mx + b, and all I've done is define m and b :)
It just takes practice to think of it the way I (and other math or science oriented people) do. Once you've spent the time studying what the symbols mean, and what the behaviors actually mean, you start to get a handle for how to translate words into formulas. I've find that it helps to think graphically... you have a starting point and a description of a behavior, and you can turn that into a graph, with which you can analyze the behavior numerically.