# ​ 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 : root2                            2.   2 : 2 : 1                                      3.   1 : 2 ; 1                                      4.    1 : 1 : 1

deepak
84 Points
5 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=  M1/2 $\sqrt{2\nu }$  / $\sqrt{Q }$ B

hence
rproton = Me1/2 $\sqrt{2\nu }$  / $\sqrt{e}$ B  where Me and e are the mass and the charges of proton

rduetron = $\sqrt{2}$Me1/2 $\sqrt{2\nu }$  / $\sqrt{e}$ B   since duetron has mass = 2Me

ralpha particle= 2 xMe1/2 $\sqrt{2\nu }$  /   $\sqrt{2}$$\sqrt{e}$ B   as mass od alpha particle is 4Me and charge =2e

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

hence option 1 is correct
Rishi Sharma
2 years ago
Dear Student,

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