A deuteron and a proton moving with equal kinetic energy enter into...

As we know, radius of circular path followed by charged particle under perpendicular magnetic field is given by,

$R = \dfrac{mv}{qB}$ and $KE = \dfrac{1}{2}m{v}^{2}$

Therefore,

⇒ R = \dfrac{\sqrt{2m\left(\text{ KE }\right)}}{qB}$Then,$R ∝ \dfrac{\sqrt{m}}{q}$</p><p>$⇒ \dfrac{{R}_{d}}{{R}_{p}} = \sqrt{\dfrac{{m}_{d}}{{m}_{p}}} \times \dfrac{{q}_{p}}{{q}_{d}} = \sqrt{\dfrac{2}{1}} \times \dfrac{1}{1} = \sqrt{2}: 1$</p><p>Thus,$x = 2.