Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1489056 • Letter: C
Question
Consider the "collision" between two helium nuclei ("alpha particles"). These particles have a mass equal to four protons and a charge equal to two protons. Two helium nuclei are heading directly towards each other with the same speed, so v1 = - v2 where v1 and v2 are the velocities of the particles. A detector determines that the speed of the particles is vi = 2.9x106 m/s when they are a distance di = 3.2x10-12 m apart. If df is the closest distance the particles approach each other, what is the ratio of the initial to final distances, di/df?
Explanation / Answer
two charges of same sign repel each other.
hence when the particles are travelling toward each other, their initial kinetic energy will be utilized in overcoming the repulsive force.
at the distance of closest approach, their speed will be zero.
mass of each particle=m=4*1.67*10^(-27) kg=6.68*10^(-27) kg
charge on each particle=charge on two protons=2*1.6*10^(-19) coloumbs
=3.2*10^(-19) coloumbs
initial total energy:
initial kinetic energy=0.5*m*v1^2+0.5*m*v2^2
=5.61788*10^(-14) J
initial potential energy=k*q1*q2/r
where k=9*10^9 F/m
q1,q2 are charges on the particles=3.2*10^(-19) C
r=distance between two particles=3.2*10^(-12) m
==>initial potential energy=9*10^9*3.2*10^(-19)*3.2*10^(-19)/(3.2*10^(-12))
=2.88*10^(-16) J
then total initial energy=5.61788*10^(-14)+2.88*10^(-16) =5.64668*10^(-14) J
let the closest distance be df.
then at this distance, speed of each particle will be zero.
hence kinetic energy =0
hence total energy=electrical potential energy
==>9*10^9*(3.2*10^(-19))^2/df=5.64668*10^(-14)
==>df=1.63211*10^(-14) m
hence di/df=3.2*10^(-12)/(1.63211*10^(-14))=196.065