A deuteron and a proton moving with equal kin | vedclass

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Question

$R =\frac{ mv }{ q _{ B }}$

$R _{ D }=\frac{\left(2 m _{ p }\right) v _{ D }}{ e B}$

$R _{ p }=\frac{\left( m _{ p }\right) v _{ p }}{ e B }$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$R =\frac{ mv }{ q _{ B }}$

$R _{ D }=\frac{\left(2 m _{ p }\right) v _{ D }}{ e B}$

$R _{ p }=\frac{\left( m _{ p }\right) v _{ p }}{ e B }$

$\frac{ R _{ D }}{ R _{ p }}=\frac{2 v _{ D }}{ v _{ p }}=\frac{2 v _{ D }}{\sqrt{2} v _{ D }}=\frac{\sqrt{2}}{1}$

$\frac{1}{2}\left(2 mp _{ D }\right) v _{ D }^{2}=\frac{1}{2} m _{ p } \cdot v _{ p }^{2}$

$\sqrt{2} v _{ D }= v _{ p }$

$x =2$

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If $r_{d}$ and $r_{p}$ are the radii of their circular paths respectively, then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt{ x }: 1$ where $x$ is ..........

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