Alpha Particle Flow: Current & Resistance

AI Thread Summary
The discussion revolves around calculating various parameters related to a beam of alpha particles with a current of 0.25 microamperes and kinetic energy of 20 MeV. To determine how many alpha particles strike a surface in 3 seconds, participants suggest using the relationship between current and charge, emphasizing unit cancellation. The number of alpha particles in a 20 cm length of the beam remains unclear to some contributors. Additionally, the potential difference required to accelerate each alpha particle to 20 MeV can be calculated using the formula V = E/q, where E is the energy and q is the charge. Overall, the thread provides insights into solving problems related to current and resistance in particle physics.
xerox2ooo
Messages
8
Reaction score
0
Can anyone give some hints on this problem?
This is about current and resistance.

Problem: A steady beam of alpha particles (q= +2e) traveling with constant kinetic energy 20MeV carries a current of 0.25 micro ampere.
(a) If the beam is directed perpendicular to a plane surfece, how many alpha particlees 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 was it necessary to accelearte each alpha particle from rest to bring it to an energy of 20 MeV?

Thanks,
 
Physics news on Phys.org
Originally posted by xerox2ooo
Problem: A steady beam of alpha particles (q= +2e) traveling with constant kinetic energy 20MeV carries a current of 0.25 micro ampere.
(a) If the beam is directed perpendicular to a plane surfece, how many alpha particlees strike the surface in 3.0s?
Remember what an ampere is? That's coulombs per second. Just start multiplying stuff until the units work out the way you want them to.

(\frac{coulomb}{second})(seconds)(\frac{\alpha}{coulomb})

coulombs and seconds will cancel out and you're left with the number of alpha particles.

(b) At any instant, how many alpha particles are there in a given 20 cm length of the beam?
This one I don't know.

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

E = Vq

V = \frac{E}{q}

E is 20MeV, q is 2e
 
Last edited:


Originally posted by ShawnD
Just start multiplying stuff until the units work out the way you want them to.
That always seems to work.
 
Thanks guys.

You guys are the best.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top