An \alpha -particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field \vec{B}, acting normal to the direction motion of the particles.

Calculate the ratio of radii of the circular paths described by them,

 

 

 

 
 
 
 
 

Answers (1)
S safeer

We already know that,

mass of proton (m_{p})=\frac{1}{4}\times\; mass \; of\; alpha\; particle (m_{\alpha })

and 

Charge on proton 

(q_{p})=\frac{1}{2}\times\; charge \; of\; alpha\; particle \; (q_{\alpha } ) 

Such that;

mv=\sqrt{2\; k\; m}, where k is kinetic energy.

the formula used :-

            r=\frac{mv}{qB}

So,     r=\frac{\sqrt{2\; k\; m}}{qB}

Hence, k, and B are the constant terms.

So,  r\propto \frac{\sqrt{m}}{q}

    So, the relation will be :-

    \frac{r_{p}}{r_{\alpha } }=\sqrt{\frac{m_{p}}{m_{\alpha }}}\frac{q_{\alpha }}{q_{p}}

           =\sqrt{\frac{m_{p}}{4m_{p }}}\frac{2q_{p }}{q_{p}}

            =\frac{1}{1}

Hence, the ratio of radii of the circular path stated by the particles is 1:1,

 

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