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the electron in the first orbit of ...

the electron in the first orbit of hydrogen atom revolves round the nucleus in a circular orbit of radius 0.5 Å . It takes `1.5xx10^(-4) ` Ps to complete one revolution . What is the centripetal force acting on the electron ?

A

`5xx10^(-5)N`

B

`6xx10^(-6)N`

C

`7xx10^(-7)N`

D

`8xx10^(-8)N`

Text Solution

Verified by Experts

The correct Answer is:
D
`T=1.5 xx10^(-4) `Pico -second
`= 1.5 xx10^(-4) xx10 ^(-12) s= 1.5 xx10^(-16) s `
`and omega =(2pi) /(T) = (2pi xx10^(16))/(1.5) rad //s`
` therefore C.P ` Force ` = m romega ^(2)`
`= (9xx10^(-31) xx0.5xx10^(-10 ) xx4 pi ^(2) xx10^(32))/(1.5xx1.5)`
`= 8 xx10^(-8) N`
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