The angular momentum of the electron mov...

The angular momentum of the electron moving around the nucleus in a circular orbit of radius `r` is given by

`m_(e)vr`

`m_(e)v//r`

`m_(e)vr^(2)`

`m_(e)v//r^(2)`

Text Solution

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The correct Answer is:

Just as linear momentum is the product of mass `(m)` and linear velocity `(V)`, angular momentum is the product of moment of inertia `(I)` and angular velocity `(omega)`. For an electron of mass `m_(e)` moving in a circular orbit of radius `r` around the nucleus,
Angular momenutm `= I xx omega`
Since `I = m_(e)r^(2)` and `omega = v//r`, where `v` is the linear velocity,
Angular momentum `= m_(e)r^(2) xx (v)/(r) = m_(e)vr`

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The orbital speed of an electron orbiting around the nucleus in a circular orbit of radius r is v. Then the magnetic dipole moment of the electron will be

ever

`(evr)/(2)`

`(ev)/(2r)`

`(vr)/(2e)`

The orbital speed of an electron orbiting around the nucleous in a circular orbit or radius r is v then the magneitc dipole moment of the electron will be

`(L_(0))/(M_(0))`

`(M_(0))/(L_(0))`

`L_(0)M_(0)`

`sqrt(M_(0))/(L_(0))`

An electron orbiting around the nucleus of an atom

has a magnetic dipole moment

exerts an electric force on the nucleus equal to that on it by the nucleus

does produce a magnetic induction at the nucleus

has a net energy inversely proportional to its distance from the nucleus.