Current from stream of particles

[jehan60188]

Homework Statement

An α (Alpha) particle consists of two neutrons and two protons bound together into a single particle with a charge of +2e. We normally measure the kinetic energy of such particles in millions "electron volts" or MeV. (1 eV = 1.6 X 10-19J).

Assume a steady stream of α particles travels with a constant energy of 20 MeV and carries a current of 2 μA.

a) If the beam is directed perpendicular to a plane target, how many α's will strike this target in 5 seconds?
b) At any given instant, how many α's are in a 15 cm length of the beam?
c) Through what electrical potential diffence [sic] is it necessary to accelerate each α from rest to bring it to the energy of 20 MeV?

Homework Equations

v = ir
r = p*l/a
E = .5*mv^2

The Attempt at a Solution

I have the velocity of the particles.
But I don't know how many particles go through a point in 1 second.

if I can get a, b falls into place.

but I have no idea how to start c

 

[jambaugh]

I have the velocity of the particles.
But I don't know how many particles go through a point in 1 second.

That is what the current gives you. 1 Ampere = 1 Coulomb/second

 

[jehan60188]

*facepalm* duh!

the current on the stream is 2 uA, so we have 2*10^-6 *1/(2e) := N1 particles going through a point
the unit is Particles/S
multiply that by 5, to get the number of particles passing through a point in 5 seconds

the velocity is about 30000000 m/s. so, it takes .15/3e7 seconds to cover 15 cm
in .15/3e7 seconds, there are (.15/3e6)*(N1) = 30287 particles going through a point

lastly, to bring a particle to V = U/q, so V = 20 MeV / (2eV) = 10000000

 

[jambaugh]

*facepalm* duh!

Happens to me all the time ;)

 

[asteriatic]

.

I can provide a response to your question. To answer part a), we can use the formula for current, I = Q/t, where Q is the charge and t is the time. In this case, we know the current is 2 μA (2 x 10^-6 A) and the time is 5 seconds. We also know that the charge of an α particle is +2e, where e is the elementary charge (1.6 x 10^-19 C). Therefore, we can calculate the number of α particles that will strike the target in 5 seconds:

I = Q/t
2 x 10^-6 = (N x 2e)/5
N = (2 x 10^-6 x 5)/2e
N = 5 x 10^-6 / 2 x 1.6 x 10^-19
N = 5 x 10^-6 / 3.2 x 10^-19
N = 1.5625 x 10^13

Therefore, approximately 1.56 x 10^13 α particles will strike the target in 5 seconds.

For part b), we can use the formula for current density, J = I/A, where I is the current and A is the cross-sectional area. We know the current is 2 μA and the area is the cross-sectional area of the beam, which we can approximate as a circle with radius r = 15 cm. Therefore, the area is A = πr^2 = π(0.15)^2 = 0.0707 cm^2. Converting this to m^2, we get A = 7.07 x 10^-5 m^2. Plugging these values into the formula for current density, we get:

J = I/A
J = (2 x 10^-6)/(7.07 x 10^-5)
J = 2 x 10^-6 / 7.07 x 10^-5
J = 2/7.07 x 10^-11
J = 2.83 x 10^5 A/m^2

Therefore, at any given instant, there are approximately 2.83 x 10^5 α particles in a 15 cm length of the beam.

For part c), we can use the formula for kinetic energy, E =