# An $\inline \alpha -particle$ and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field $\inline \vec{B},$ acting normal to the direction motion of the particles.Calculate the ratio of radii of the circular paths described by them,

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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