Question

A proton, a deutron and an alpha particle, after being accelerated through the same potential difference enter a region of uniform magnetic field vecB, in a direction perpendicular to vecB. Find the ratio of their kinetic energies. If the radius of proton's circular path is 7cm, what will be the radii of the paths of deutron and alpha particle?

Answer 1

Answer: hope this will help u

Explanation:

1:1:2, rd=72–√cm, rα=72–√cm

Solution :

Charge on proton =e, charge on deutron =e,

Charge on α-particle =2e.

As qV=12mv2=K (kinetic energy)

So, K∝q

∴Kp:Kd:Kα=qp:qd:qα=e:e:2e

=1:1:2

Radius of circular path of a charged particle in a magnetic field is

r=mvqB=2mK−−−−√qB

(∵K=12mv2 or mv=2mK−−−−√)

For proton, rp=2mpKp−−−−−−√eB=7cm

For deutron, rd=2mdKd−−−−−−√eB

=2×2mp×KpeB−−−−−−−−−−−−√=2–√rp=2–√×7cm

For α-particle, rα=2mαKα−−−−−−√2eB

=2×4mp×2Kp2eB−−−−−−−−−−−−−√=2–√rp=2–√×7cm