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A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speedand maximum accelerationof each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

Short Answer

Expert verified

The maximum speed of proton is 4.43×104m/s

The maximum acceleration of proton is 5.44×1018m/s2

The maximum speed of alpha particle is 1.11×104m/s

The maximum acceleration of the alpha particle is 1.36×1018m/s2

The maximum acceleration occurs immediately after the release. It takes a longer time to attain the maximum speed.

Step by step solution

01

(a) Determination of the maximum acceleration of proton and the alpha particle.

The proton charge is qp=+e

The proton mass is mp=1.67×1027kg

Right after the release, the force is maximum and therefore the acceleration is also maximum,

Acceleration ap=Fm0

Now, F=kqqr2

And r is the distance which is equal to0.225 nm.

So, solve for F,

F=8.99×109N×m2/C2(2)1.60×1019C20.225×109m2=9.09×109N

ap=9.09×109N1.67×1027kg=5.44×1018m/s2

Similarly, for the alpha particle,

The charge is qα=+2e

The mass is mα=4mp=6.68×1027kg

Acceleration aα=Fmα

aα=9.09×109N6.68×1027kg=1.36×1018m/s2

Thus the acceleration of the proton and alpha particle respectively is 5.44×1018m/s2 and 1.36×1018m/s2.

02

(b) Determination of the maximum speed of the alpha particle and the proton.

Speed is calculated by the law of conservation of energy,

U1+K1=U2+K2

Now, K1 = 0 and U2 = 0.

So, U1 = K2

Solve for U1,

localid="1668229301464" U1=8.99×109N×m2/C2(2)1.60×1019C0.225×109m=2.05×1018J

So this is the total kinetic energy of the system, localid="1668229340261" K2=12mpvp2+12mαvα2=2.05×1018J

This amount is divided between the proton and the alpha particle, Therefore, use the momentum conservation principle to solve for the speed of the proton.

p1=p2mpvpmαvα=0va=mpmαvp

So, kinetic energy is,

K2=12mpvp2+12mαvα2=12mpvp21+mpmα

Solve for vp,

localid="1668229807440" Vp=2K2mp1+mpma=22.05×1018J1.67×1027kg1+14=4.43×104m/s

Similarly, solve for vα,

localid="1668229670474" va=mpmavp=12124.43×104m/s=1.11×104m/s

Thus the speed of the proton and alpha particle respectively is localid="1668229814402" 4.43×104m/sand 1.11×104m/s.

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