Answer
$ l=2.49\times10^{74}$
Work Step by Step
The magnitude of the orbital angular momentum$(\vec L)$ of an electron trapped in an atom is given by
$L=\sqrt {l(l+1)}\hbar,\;\;\;\text{for}\;l=0,1,2,...,(n-1)$
The average angular momentum of the earth is
$L=mvr$,
where we have treated the Earth as a point mass.
The mean distance from Earth to Sun is $r=1.496×10^{11}\;m$.
Mass of the Earth is $m=5.974×10^{24}\;kg$
Earth takes $t=365\;days$ to complete one revolution around the Sun.
Speed of the earth$(v)=\frac{2\times\pi\times r}{t}$
$v=\frac{2\times\pi\times 1.496×10^{11}}{365\times 24\times 3600}\;m/s=2.98\times10^{4}\;m/s$
Therefore, Earth's angular momentum is
$L=5.974\times10^{24}\times 2.98\times10^{4}\times \times1.496×10^{11}=2.663\times10^{40}\;kg.m^2s^{-1}$
For large value of $L$, we can write
$\sqrt {l(l+1)}\hbar\approx l\hbar$
Therefore for earth:
$l\hbar=2.663\times10^{40}$
or, $ l=2.49\times10^{74}$
