The UCA Physics department used to maintain a 3 million Volt particle accelerato

Question

The UCA Physics department used to maintain a 3 million Volt particle accelerator. It was used to accelerate Helium nuclei for nuclear physics experiments. The Helium was doubly ionized, He^+2, and had a mass of 6.68 times 10^- 27 kg. Answer the following: How fast were the helium nuclei moving if they started from rest after traveling through this potential difference? How close to a gold nucleus with charge +79e could the Helium nucleus approach? The radius of a gold nucleus is about 7.3 times 10^- m. What voltage would the UCA accelerator need to bring the He^+2 nucleus this close to the gold nucleus? Again neglect any recoil of the gold nucleus.^2-

Explanation / Answer

part a:

electrical potential energy imparted on the He ion after travelling through the voltage difference

=charge*potential difference

=2*1.6*10^(-19)*3*10^6 J

=9.6*10^(-13) J

this potential energy gets converted to kinetic energy of the ion.

if speed of the ion is v m/s,

then 0.5*mass*speed^2=kinetic energy

==>0.5*6.68*10^(-27)*speed^2=9.6*10^(-13)

==>speed=sqrt(9.6*10^(-13)/(0.5*6.68*10^(-27))=1.6954*10^7 m/s

part b:

assuming at the closest distance, electrical potential energy=initial kinetic energy

==>k*q1*q2/d=initial kinetic energy

where k=coloumb's constant

q1=charge on helium ion

q2=charge on gold nucleus

d=distance between the two charges

==>d=9*10^9*2*1.6*10^(-19)*79*1.6*10^(-19)/(9.6*10^(-13))

=3.792*10^(-14) m

part c:

let voltage required be V volts,

then 9*10^9*2*1.6*10^(-19)*79*1.6*10^(-19)/(7.3*10^(-15))=2*1.6*10^(-19)*V

==>V=(9*10^9*2*1.6*10^(-19)*79*1.6*10^(-19)/(7.3*10^(-15)))/(3.2*10^(-19))

=1.5584*10^7 volts

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