A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference ΔV. Each of the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius rp. (a) In terms of rp, determine the radius rd of the circular orbit for the deuteron. rd = (b) In terms of rp, determine the radius rα for the alpha particle. rα =

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gatorchia gatorchia · Answer 1 · 2019-11-07T13:05:36+01:00

Answer: a) (2)^1/2*rp; b) ra= (2)^1/2*rp

Explanation: In order to explain this problem we have to consirer a circular motion for charged particles with velocity in a magnetic field.

The we know that:

the Ek=q*V=1/2*m*v^2 then

v=(2*q*V/m)^1/2

The magnetic force when v is perpendicular to B, is equal

Fm=q*v*B

when the charged particle is moving a circular motion, we have:

Fm=m*ac where ac is the centripetal acceleration which is equal,

ac=v^2/R where R is the radius of the circular trajectory.

Finally we obtain;

q*v*B=m*v^2/R

R=m*v/(q*B) using that v= (2*q*V/m)^1/2

we have; R=(1/B)*(2*V*m/q)^1/2

For a proton we have

rp=(1/B)* (2*V*mp/e)^1/2

For a deuteron ( charge +e and mass 2mp)

rd=(1/B)* (2*V*2*mp/e)^1/2=(1/B)*(4*V*mp/e)^1/2=(2)^1/2*rp

For a alpha particle ( charge +2e, mass 4mp) =

ra=(1/B)*(2*V*4*mp/2*e)^1/2=(1/B)*(4*V*mp/e)^1/2=(2)^1/2*rp