# A proton, a deuteron and an alpha particle, are accelerated through the same potential difference and then subjected to a uniform magnetic field $\vec{B}$ , perpendicular to the direction of their motions. Compare (i) their kinetic energies, and (ii) if the radius of the circular path described by proton is 5 cm, determine the radii of the paths described by deuteron and alpha particle.

(i) We know that,

Kinetic energy (K.E

$=\frac{1}{2}mv^{2}$

$K.E=qV$

The kinetic energy of the proton is given as $\Rightarrow \frac{1}{2}m_{p}v^{2}_{1}=qV$

and kinetic energy of deuteron is given as $\Rightarrow \frac{1}{2}m_{d}v^{2}_{2}=qV$

and kinetic energy of $\alpha$ - particle is given as$\Rightarrow \frac{1}{2}m_{\alpha }v^{2}_{3}=2qV$

Hence, $(K.E)_{p}:(K.E)_{d}:(K.E)\alpha :qV:qV:2qV$

therefore, $1:1:2$

(iii) We know that,

$F=qvB$

Or      $F=\frac{mv^{2}}{r}$

where, q = charge                    ,    v=velocity

B = magnetic field         ,    m=mass

So, of compairing both forces, we get

$qvB=\frac{mv^{2}}{r}$

$\because \; qB=\frac{mv}{r}$

$\therefore \; r=\frac{mv}{qB}$

Such that, given ; r = 5cm,

So, $5=\frac{mv}{qB}=\frac{\sqrt{2m(KE})}{qB}$

If the mass of a proton is m then the mass of deuteron is 2m and the mass of alpha particle is 4m

Hence,

$r_{1}:r_{2}:r_{3}=5:5\sqrt{2}:5\sqrt{2}$

$=1:\sqrt{2}:\sqrt{2}$

### Preparation Products

##### Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
##### Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
##### Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-