A steady beam of alpha particles (q = + 2e, mass m = 6.68 x 10^27 kg) traveling

Question

A steady beam of alpha particles (q = + 2e, mass m = 6.68 x 10^27 kg) traveling with constant kinetic energy 21 MeV carries a current of 0.30 mu A. If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in 4.1 s At any instant, how many alpha particles are there in a given 20 cm length of the beam Through what potential difference in volts is it necessary to accelerate each alpha particle from rest to bring it to an energy of 21 MeV Number Units Number Units Number Units

Explanation / Answer

a) Correct Answer

b) current = 0.3 mu A = 0.3 mu C/ sec

length = 20 cm

kinetic energy = 21 MeV

hence velocity, v = sqrt(2 KE/m) = sqrt(2*21*10^6*1.6*10^-19/(6.68*10^-27)) = 3.17*10^7 m/s

time taken = 0.2/(3.17*10^7) sec

hence total charge = 0.3*10^-6*0.2/(3.17*10^7) = 1.89*10^-15 C

Hence number of particles = 1.89*10^-15/(2*1.6*10^-19) = 5906

c) Suppose the potential difference is V then the achieved kinetic energy = 2eV

hence 2eV = 21 MeV

Hence V = 10.5 MV = 1.05*10^7 Volts.

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