An electron of mass `9 xx 10^(-31)` kg revolves around a nucleus in a circular orbit of radius `0.53 Å`. What is the angular momentum of the electron? (Velocity of electron is, `v=2.2 xx 10^(6) ms^(-1)).`

Explanation

Mass of the electron, `m=9xx10^(-31)kg`
Radius of the electron, `r=0.53 Åxx10^(-10)m`
Velocity of the electron, `v=2.2xx10^(6)ms^(-1)`
ngular momentum of electronis, `L= I omega`
Electron is considered as a point mass. Hence, its moment of inertia is, `I -mr^(2)`
The relation, `omega=(v)/(r)` could be used.
Angular momentum,`L=mr^(2)xx(v)/(r)=mvr`
`=9.1xx10^(-31)xx2.2xx10^(6)xx0.53xx10^(-10)`
`L=1.06xx10^(-34) kg m^(2)s^(-1)`
`=9.1xx10^(31)xx2.2xx10^(6)xx0.53xx10^(-10)`
`L=1.06xx10^(-34) kg m^(2)s^(-1)`