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Angular momentumis quantised i.e it us an integral multiple of h/2 according to de-broglie, electrons have dual nature of behaviour and behaves like particle and wave.

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Hello,

When an electron of mass m is orbiting in a circular orbit of radius r with speed v, its angular momentum is given by

L = m * v * r (Equation 1)

According to de-broglie, electrons have dual nature of behaviour and behaves like particle and wave.

If electron has momentum m * v, then its wavelength lambda is given by

lambda = h / (m * v) (Equation 2 ) where h is plancks constant.

When Bohr developed his theory of hydrogen atom, it was stated that electrons revolving in stationary orbit will never loose energy and this stationary orbit satisfies the condition that the circumference of orbit is integral multiples of wavelength.

Hence 2 * pie * r = n * lambda     ( Equation 3)

if we combine all equations (1), (2) and (3),  we get angular momentum, L:

(h / lambda) * (n * lambda) /2 * pie = n * ( h / 2 * pie)

where n is called orbital quantum number. For ground state n = 1, first excited state n = 2 and so on

Hence quantisation of angular momentum is integral multple of h / 2 * pie .

Basically, it means that The quantity h / 2 * pie is multiplied by integers and no fraction is involved.

Also, Orbital quantum number n  depends upon whether the electron is in ground state or excited state.

Hope this helps!