An electron of mass 9×10−31kg moves around a nucleus in a circular orbit of radius 2A0 under the action of centripetal force 3.2N then the equivalent electric current in ampere is

i=qT=qv2πrF=mv2r v=Frmi=q2πrFrmsubstituting the values we get i=323π A

An electron of mass 9×10−31kg moves around a nucleus in a circular orbit of radius 2A0 under the action of centripetal force 3.2N then the equivalent electric current in ampere is

i=qT=qv2πrF=mv2r v=Frmi=q2πrFrmsubstituting the values we get i=323π A

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An electron of mass $9 \times 10^{- 31} kg$ moves around a nucleus in a circular orbit of radius 2A⁰ under the action of centripetal force 3.2N then the equivalent electric current in ampere is

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$\frac{3}{32 π}$

$\frac{16}{3 π}$

$\frac{3}{16 π}$

$\frac{32}{3 π}$

answer is D.

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Detailed Solution

$i = \frac{q}{T} = \frac{qv}{2 πr}$

$F = \frac{m v^{2}}{r} v = \sqrt{\frac{F r}{m}}$

$i = \frac{q}{2 πr} \sqrt{\frac{Fr}{m}}$

substituting the values we get

$i = \frac{32}{3 π} A$

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An electron of mass 9×10−31kg moves around a nucleus in a circular orbit of radius 2A0 under the action of centripetal force 3.2N then the equivalent electric current in ampere is

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An electron of mass $9 \times 10^{- 31} kg$ moves around a nucleus in a circular orbit of radius 2A⁰ under the action of centripetal force 3.2N then the equivalent electric current in ampere is