Magnetism> Q. A proton , deutron and alpha particl...

Q.A proton , deutron and alpha particle are accelerated by same potantial , enters in a uniform magnetic field perpendicularly.Ratio of radii of circular paths respectively is-1.1: root2 : root22.2 : 2 : 13.1 : 2 ; 14.1 : 1 : 1

Naina Singh , 8 Years ago

Grade 12th pass

2 Answers

deepak

Last Activity: 7 Years ago

the radius of the trajectory of the particle is given by

r= MV / BQ where M is the mass of the proton V is its velocity in the field B is the magnetic field intensity Q is the charge

V = $\sqrt{2\nu Q}/\sqrt{M}$ here $\nu$ is the accerating potential

therefor,

r= M^1/2 $\sqrt{2\nu }$ / $\sqrt{Q }$ B

hence

r_proton = M_e^1/2 $\sqrt{2\nu }$ / $\sqrt{e}$ B where M_e and e are the mass and the charges of proton

r_{duetron =} $\sqrt{2}$ M_e^1/2 $\sqrt{2\nu }$ / $\sqrt{e}$ B since duetron has mass = 2M_e

r_{alpha particle}= 2 xM_e^1/2 $\sqrt{2\nu }$ / $\sqrt{2}$ $\sqrt{e}$ B as mass od alpha particle is 4M_e and charge =2e

hence the ratio is 1 : $\sqrt{2}$ : $\sqrt{2}$

hence option 1 is correct

hope this is helpful XD

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

Radius of circular path r = mv/Bq
Using p = mv = √2mK
where K is the kinetic energy of the particle.
We get r = (√2mK)/Bq
⟹ r ∝ √m/q
Given : qα = 2qp and qd = 2qp where qp is the charge of proton.
Also, mα = 4mp and md = 2mp where mp is the mass of proton.
∴ rp : rd : rα = √mp/qp : √md/qd : √mα/qa
Or rp : rd : rα = √mp/qp : √2mp/2qp : √4mp/2qp
⟹rp : rd : rα = √2 : 1 : √2

Thanks and Regards