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.

 

 

Answers (1)
S safeer

(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}

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