Two protons are moving toward each other in what will be a head-on collision. Ea

Question

Two protons are moving toward each other in what will be a head-on collision. Each has an initial kinetic energy Ki when they are far apart.

(a) Find the value of Ki that allows the protons to get to a center-to-center separation of

1.2*10^ (-15) m.
(b) The average kinetic energy of particles in a gas at absolute temperature T is Kav = (2/3) kB T , where kB is Boltzmann’s constant. Find the temperature at which Ki = 10Kav (At this temperature, about one in a million protons has a kinetic energy of Ki or greater.)

Explanation / Answer

a) we will use conservation of energy

Initial KE = Final pE

2 ( kE ) = K ( q^2) /r

2KE = 9 x 10^9 ( 1.6 x 10^-19)^2/ ( 1.2 x 10^-15 )

KE =9.6 x 10^ -14 j

b) 9.6 x 10^ -14 = 10 ( 2/3 ) ( T) ( 1.38 x 10^-23)

T( in kelvin) = 1.04 x 10^ 9 K

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