Modern Physics> This question is from mixture of modern p...

This question is from mixture of modern physics and magnetism. Ans is10(2)^1/2

Psd , 7 Years ago

Grade 12th pass

1 Answers

Khimraj

Last Activity: 7 Years ago

radius of orbit in magnetic field is given by

r = mv/qB

Let potential difference is V

kinetic energy E = (½)mv² = qV

so mv = $\sqrt{2mE}$ = $\sqrt{2mqV}$

So r = $(1/B)\sqrt{2mV/q}$ $\propto \sqrt{m/q}$

m/q for proton = 1/e

and for $\alpha$ particle = 4/2e = 2/e

so radius for $\alpha$ particle = $\sqrt{2}$ times pf proton = 10 $\sqrt{2}$ cm.

Hope it clears. If you like answer then please aapprove it.

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Modern Physics> This question is from mixture of modern p...

This question is from mixture of modern physics and magnetism. Ans is10(2)^1/2

Psd , 7 Years ago

Khimraj

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