# Thread: Alpha Particles? (Don't really know a good title for this)

1. ## Alpha Particles? (Don't really know a good title for this)

A steady beam of alpha particles (q = +2e) traveling with constant kinetic energy 20 MeV carries a current of 0.25 $\displaystyle \mu A$.

(a) If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in 3.0s?

(b) At any instant, how many alpha particles are there in a given 20 cm length of the beam?

(c) Through what potential difference is it necessary to accelerate each alpha particle from rest to bring it to an energy of 20 MeV?

2. Originally Posted by Aryth
A steady beam of alpha particles (q = +2e) traveling with constant kinetic energy 20 MeV carries a current of 0.25 $\displaystyle \mu A$.

(a) If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in 3.0s?

(b) At any instant, how many alpha particles are there in a given 20 cm length of the beam?

(c) Through what potential difference is it necessary to accelerate each alpha particle from rest to bring it to an energy of 20 MeV?
(a) Note that Amp = C/sec ..... So convert q into Coulomb and divide that answer into 0.25 x 10^-6. Then multiply by 3.

(b) Calculate the speed of an alpha-particle and hence calculate the time it takes to travel 20 cm. Then calculate how many alpha particles strike the surface in that time.

(c) Is it over a particular distance?

3. (a) I got it, thanks

(b) We're studying circuits, Loops, Ohm's Law and all that... I don't have a formula or anything for $\displaystyle \alpha$-particles...

(c) It has no distance measurement. That's all it gives us.

4. Originally Posted by Aryth
(a) I got it, thanks

(b) We're studying circuits, Loops, Ohm's Law and all that... I don't have a formula or anything for $\displaystyle \alpha$-particles... Mr F says: Get the speed using the known K.E.

(c) It has no distance measurement. That's all it gives us.
I'll try to post on (c) later as I have no time now.

5. Awesome. I got b, thanks again. I have been pounding at c and can't figure it out...

6. Originally Posted by Aryth
Awesome. I got b, thanks again. I have been pounding at c and can't figure it out...
Consider this:

$\displaystyle E=Vq$

$\displaystyle V=\frac{E}{q}$

$\displaystyle E=20MeV$ and $\displaystyle q=2e$