A proton and an alpha particle are projected in a magnetic field which exists in the width of region d. Compare the angles of deviation suffered by the proton and the alpha particle if before entering the magnetic field both the particles.
a. have the same momentum,
b. have the same kinetic energy, and
c. are accelerated through the same potential difference. Take `m_alpha = 4 m_p, q_alpha = 2 q_p.`

Explanation

The radius of circulation is `R=(mv)/(Bq)=p/(Bq)=(sqrt(2mK))/(Bq)`
As given, angle of deviation is `theta`, it is the angle between directions
of final velocity and intial velocity.

In `DeltaOAB, sin theta=d/R=(dBq)/(mv)=(dBq)/p=(dbq)/(sqrt(2mK))`
(a) `p_(alpha)=p_p, (sin theta_p)/(sin theta_(alpha))=(q_p)/(q_(alpha))=1/2`
(b) `K_(alpha)=K_P,hence (sin theta_p)/(sin theta_(alpha)) =(q_p)/(q_(alpha))xx sqrt(m_(alpha)/(m_(p)))`
`=1/2xxsqrt(4/1)=1`
(c) When a charged particle is acclerated throug a potential
difference V, gain of kinetic energy is qV
`K_p=q_(alpha)V, K_(alpha)=q_(alpha)V`
Now, `sin theta=(dBq)/(sqrt(2mqV))=(dBsqrtq)/(sqrt(2mv))`
Hence, `(sin theta_p)/(sin theta_(alpha))=sqrt((q_p)/(q_(alpha)))xxsqrt((m_alpha)/(m_p))=sqrt(1/2)xxsqrt(4/1)=sqrt2`.