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Standard XII

Physics

Circular Kinematics

Ionized hydro...

Question

Ionized hydrogen atoms and $α$ -particles with same momenta enters perpendicular to a constant magnetic field $B$ . The ratio of radii of their paths i.e. $r_{H} : r_{α}$ will be

2 : 1

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1 : 2

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4 : 1

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Solution

The correct option is A 2 : 1
$\frac{q_{H}}{q_{α}} = \frac{1}{2} r = \frac{m v}{q B}$
For same momenta, $r \propto \frac{1}{q}$
$\frac{r_{H}}{r_{α}} = \frac{q_{α}}{q_{H}} = \frac{2}{1}$

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Ionized hydrogen atoms and α-particles with same momenta enters perpendicular to a constant magnetic field B. The ratio of radii of their paths i.e. rH:rα will be

The correct option is A 2 : 1 qHqα=12 r = mvqB For same momenta, r∝1q rHrα=qαqH=21

Ionized hydrogen atoms and $α$ -particles with same momenta enters perpendicular to a constant magnetic field $B$ . The ratio of radii of their paths i.e. $r_{H} : r_{α}$ will be

The correct option is A 2 : 1
$\frac{q_{H}}{q_{α}} = \frac{1}{2} r = \frac{m v}{q B}$
For same momenta, $r \propto \frac{1}{q}$
$\frac{r_{H}}{r_{α}} = \frac{q_{α}}{q_{H}} = \frac{2}{1}$