Question

# The orbital angular momentum of an electron in 2s orbital is:(a) $\cfrac { h }{ 4\pi }$(b) zero(c) $\cfrac { h }{ 2\pi }$(d) $\sqrt { 2 } \cfrac { h }{ 2\pi }$

Hint: According to the quantum mechanical model, the total angular momentum of an electron comprises its spin angular momentum and the orbital angular momentum. The orbital angular momentum of an electron is its angular momentum when it is revolving in its orbital such that its spin is not taken into consideration.

According to classical physics the angular momentum is the vector product of the position vector and the linear momentum vector.
$L=r\times p$
The angular momentum of an electron according to the Bohr’s model is equal to mvr or $\cfrac { nh }{ 2\pi }$ (where m is the mass of the electron, v is the velocity of the electron, r is the radius of the orbit, n is the orbit of the electron and h is Planck’s constant).
According to Bohr, the angular momentum of an electron in an orbit is quantized and is an integral multiple of $\cfrac { h }{ 2\pi }$. This can be understood by De-Broglie’s equation. According to De-Broglie, the electron behaves like a wave and hence can give rise to standing waves while moving in its orbit. Bow since an orbit is circular, hence the standing waves can only be formed if the circumference of the orbit in an integral multiple of the full wavelength i.e.
$2\pi { r }_{ k }=k\lambda$ (where ${ r }_{ k }$ is the radius of the kth orbit, k=1,2,3…)
But, the De-Broglie wavelength is given by:
$\lambda =\cfrac { h }{ p }$ (where p is the linear momentum and h is Planck’s constant)
Since linear momentum is the product of mass and velocity,
$\Rightarrow \lambda =\cfrac { h }{ m{ v }_{ k } }$ (where ${ v }_{ k }$ is the velocity of the electron in the kth orbit)
Hence, $m{ v }_{ k }{ r }_{ k }=\cfrac { kh }{ 2\pi }$
Now, the Bohr’s model is not followed; rather the quantum-mechanical model is followed. According to this model, the total angular momentum of an electron consists of its spin angular momentum and the orbital angular momentum. The orbital angular momentum of an electron is its angular momentum when it is revolving in its orbital such that its spin is not taken into consideration. It can be calculated using the formula:
$orbital\quad angular\quad momentum=\sqrt { l(l+1) } \cfrac { h }{ 2\pi }$
Where l is the azimuthal quantum number of the electron. For the 2s orbital, the value of l is zero. Hence the value of the orbital angular momentum will be zero.

Therefore the correct answer is (b) zero.

Note:
The formula for the orbital angular momentum can be derived using the classical formula of the angular momentum. For that we would have to write the expression for vector L in terms of its spherical polar coordinates. Then we would have to solve the Eigenvalue equation for ${ L }^{ 2 }$ vector from where we will get the Eigenvalues for $\theta$ and by taking its square root we will get the equation for the orbital angular momentum.