A proton and an alpha particle moving with same kinetic energy enter in the region of uniform magnetic field perpendicular to it. The ratio of radii of their trajectories will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 : 1
Using Lorentz force,
$qvB=\frac{mv^2}{r}$ [Moving in circle]
$⇒r=\frac{mv}{qB}=\frac{m}{qB}\sqrt{\frac{2KE}{m}}$ $[v=\sqrt{\frac{2KE}{m}}]$
$=\frac{\sqrt{2mKE}}{qB}$
Mass of proton = $m_p$
Charge of proton = $e$
Mass of Alpha particle = $4m_p$
Charge of Alpha particle = $2e$
$⇒\frac{r_α}{r_p}=\frac{\frac{\sqrt{(4m_p)KE}}{2eB}}{\frac{\sqrt{2m_pKE}}{eB}}$
$=\frac{2}{2}=1$