A proton and deuteron ion having the same kinetic energies enter a region of uniform magnetic field perpendicularly. Deuteron's mass is twice that of proton. Calculate the ratio of the radii of their circular paths.

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Solution

Radius of circular path $(r) = \frac{m v}{q B}$
$m \to$ mass
$v \to$ velocity
$q \to$ charge
$B \to$ magnetic field.
$K E = \frac{1}{2} {m v}^{2}$
Let, $m_{p} = m$ , then $m_{d} = 2 m$
${(K E)}_{p} = {(K E)}_{d}$
$\frac{1}{2} m_{p} v_{p}^{2} = \frac{1}{2} m_{d} v_{d}^{2}$
$m . v_{p}^{2} = 2 m v_{d}^{2}$
$v_{p} = \sqrt{2} v_{d}$
$\frac{r_{p}}{r_{d}} = \frac{m v_{p} \times e . B}{e . B .2 m . v_{d}} = \frac{\sqrt{2} v_{d}}{2 v_{d}} = \frac{1}{\sqrt{2}}$
$\frac{r_{d}}{r_{p}} = \sqrt{2}$

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A proton and deuteron ion having the same kinetic energies enter a region of uniform magnetic field perpendicularly. Deuteron's mass is twice that of proton. Calculate the ratio of the radii of their circular paths.

Radius of circular path (r)=mvqBm→ massv→ velocityq→ chargeB→ magnetic field.KE=12mv2Let, mp=m, then md=2m(KE)p=(KE)d12mpv2p=12mdv2dm.v2p=2mv2dvp=√2vdrprd=mvp×e.Be.B.2m.vd=√2vd2vd=1√2rdrp=√2