MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        
A proton and a deuteron with same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field.the ratio of the radii of the circular trajectories described by them is
10 months ago

Answers : (2)

Vikas TU
11138 Points
							
Hiii 
For Proton Let r(p) =x cm 
For deutron Let r(d)  = [2md(k)d]0.5/eB  = [4mp(k)p]0.50  = 1.41x
For alpha particles r(a) = [2ma(k)a]0.50/2eB = 2x cm 
So the ratio will be 1:sqrt2:2 or 1:1:sqrt2
10 months ago
Khimraj
3008 Points
							

Radius of trajectory is given by , r= MV / BQ

Similarly , V =

\sqrt{2vQ/M}

substitute and you will get ,

r= sqrt ( 2MV / Q)B ………………….. (1)

Here, V and B are constants … so we need to look into the mass and charge of each element.

  1. Proton mass is 1 a.m.u and charge is +1e
  2. For deutron , mass = 2 a.m.u
  3. Alpha partical has mass 4 a.m.u and charge 2e

Substituting these values in (1) you get the ratios as 1 : sqrt(2) : sqrt (2)

 
9 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details