A proton, a deutron and an alpha particle, after being accelerated through the same potential difference enter a region of uniform magnetic field `vecB`, in a direction perpendicular to `vecB`. Find the ratio of their kinetic energies. If the radius of proton's circular path is `7cm`, what will be the radii of the paths of deutron and alpha particle?

Answer

`1:1:2`, `r_d=7sqrt2cm`, `r_alpha=7sqrt2cm`

Explanation

Charge on proton `=e`, charge on deutron =e,
Charge on `alpha`-particle `=2e`.
As `qV=1/2mv^2=K` (kinetic energy)
So, `Kpropq`
`:. K_p: K_d: K_alpha=q_p: q_d: q_alpha=e : e : 2e`
`=1:1:2`
Radius of circular path of a charged particle in a magnetic field is
`r=(mv)/(qB)=(sqrt(2mK))/(qB)`
(`:'K=1/2mv^2` or `mv=sqrt(2mK)`)
For proton, `r_p=(sqrt(2m_pK_p))/(eB)=7cm`
For deutron, `r_d=(sqrt(2m_dK_d))/(eB)`
`=sqrt((2xx2m_pxxK_p)/(eB))=sqrt2r_p=sqrt2xx7cm`
For `alpha`-particle, `r_alpha=(sqrt(2m_alphaK_alpha))/(2eB)`
`=sqrt((2xx4m_pxx2K_p)/(2eB))=sqrt2r_p=sqrt2xx7cm`