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 particales 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,

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 particales 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,

Answer to A:

1 Coulomb = 6.241509...x 10^18 electron charges

1 Ampere = 1 Coulomb / Sec

Thus:

0.25x10^-6 Ampere = 0.25x10^-6 Coulombs/sec

3sec * 0.25*10^-6 = 0.75x10^-6 coulombs

0.75x10^-6 Coulombs * 6.24x10^18 electron charges / coulomb = 4.68x10^12 electron charges

Particals = 4.68*10^12 +e / +2e / particle =

1 Coulomb = 6.241509...x 10^18 electron charges

1 Ampere = 1 Coulomb / Sec

Thus:

0.25x10^-6 Ampere = 0.25x10^-6 Coulombs/sec

3sec * 0.25*10^-6 = 0.75x10^-6 coulombs

0.75x10^-6 Coulombs * 6.24x10^18 electron charges / coulomb = 4.68x10^12 electron charges

Particals = 4.68*10^12 +e / +2e / particle =

**2.36*10^12 alpha particles in 3 seconds**An alpha particle is 6.64465598 ? 10-27 kg in mass

1eV = 1.6*10^-19 Joules

E= 1.6*10^-19 * 20*10^6 = 32 * 10^ -13 Joules

E = 0.5mv^2

v = sqrt( 2E/m ) = sqrt( 64*10^-13 Joules/6.64465598 ? 10-27 kg )

v = 3.103*10^7 m/s

Particles / sec = 2.36*10^12 / 3 = 0.7866 *10 ^12 Particles / sec

Thus Particals per meter = 0.7866*10^12 P/s / 3.103*10^7 m/s = 0.2535 * 10^5 particles / meter

0.20 meters * 0.2535 * 10^5 particles / meter =

1eV = 1.6*10^-19 Joules

E= 1.6*10^-19 * 20*10^6 = 32 * 10^ -13 Joules

E = 0.5mv^2

v = sqrt( 2E/m ) = sqrt( 64*10^-13 Joules/6.64465598 ? 10-27 kg )

v = 3.103*10^7 m/s

Particles / sec = 2.36*10^12 / 3 = 0.7866 *10 ^12 Particles / sec

Thus Particals per meter = 0.7866*10^12 P/s / 3.103*10^7 m/s = 0.2535 * 10^5 particles / meter

0.20 meters * 0.2535 * 10^5 particles / meter =

**5.07*10^3 Particles**1eV is the energy an electron gains through a potential difference of 1 V.

An alpha particle is equivalent to 2e. Thus the energy gained by an alpha particle through a potential difference of 1V is 2eV.

Thus to get alpha particles at 20MeV,

An alpha particle is equivalent to 2e. Thus the energy gained by an alpha particle through a potential difference of 1V is 2eV.

Thus to get alpha particles at 20MeV,

**you need to have a potential difference of 10MV. (10*10^6 Volts).**<head explodes>

LoL :grin:

Me + Google = Home grown physicist

( I have no idea if im right, i just know math :grin: )

Me + Google = Home grown physicist

( I have no idea if im right, i just know math :grin: )

Hi thanks guys.

I solved the problem successfully and by the way, I also posted this question on

www.physicsforums.com and if you would like to see another hints here is the link

http://physicsforums.com/showthread.php?t=16937

By the way, this site has pretty interesting discussion so if you are interested in physics it might be useful.

I solved the problem successfully and by the way, I also posted this question on

www.physicsforums.com and if you would like to see another hints here is the link

http://physicsforums.com/showthread.php?t=16937

By the way, this site has pretty interesting discussion so if you are interested in physics it might be useful.

I skimmed through your links, and didnt see a direct answer for A or B. I know C is correct, but im currious how far off the mark i was with A & B. The presented solution equation to A is what i followed in principal, but am currious what the numerical values are....

Regards,

:NaN:

Regards,

:NaN: