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An electron in the hydrogen atom circles...

An electron in the hydrogen atom circles around the proton with a speed of `2.18xx10^(6)`m/s in an orbit of radius `0.53Å` . The equivalent current is

A

`1.048xx10^(-2)A`

B

`1.048xx10^(-3)A`

C

`1.048xx10^(-4)A`

D

`1.058xx10^(-4)A`

Text Solution

Verified by Experts

The correct Answer is:
B
`I=q/t=e/t=(e v)/(2 pi r)`
`= (1.6xx10^(-19)xx2.18xx10^(6))/(6.28xx0.53xx10^(-10))`
`=1.048 xx 10^(-3)A`
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