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​ 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

​ 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

Grade:12th pass

2 Answers

deepak
84 Points
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=  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
hope this is helpful XD
Rishi Sharma
askIITians Faculty 646 Points
3 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

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