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An electron of mass 9xx10^(-31) kg revol...

An electron of mass `9xx10^(-31)` kg revolves in a circle of radius 0.53 Å around the nucleus of hydrogen atom with a velocity of `2.2xx10^(6) "ms"^(-1)`. What is the angular momentum of the electron ?

A

`(h)/(2pi)`

B

`(3h)/(3pi)`

C

`(h)/(pi)`

D

`(h)/(3pi)`

Text Solution

Verified by Experts

The correct Answer is:
A
Electron is moving in the first orbit
n=1
Angular momentum = `(nh)/(2pi)`
`:. For " " n = 1, L = (h)/(2pi)`
`[L = mvr = 9xx10^(-31)xx22xx10^(6)xx0.53xx10^(-10)`
`=1.0494xx10^(-34)"j.s"`
`"Also" " " (h)/(2pi) = (6.6xx10^(-34))/(2xx3.14)`
`L = (h)/(2pi)]`
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