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)

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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