The correct option is D 2hπ
Given orbital angular momentum = √3 hπ
We know that orbital anugular momentum of an orbital is given by √l(l+1) h2π
On equating both we get,√l(l+1) h2π = √3 hπ
From here, l(l+1) = 12
∴ l= −4 , 3 (But -4 is not allowed as value of l cannot be negative).
Hence, l = 3 is final accepted value.
Hence possible values of n = 4 , 5 , 6
We also know that according to Bohr's model angular momentum for an orbit is given by mvr=nh2π
Thus, possible angular momentum values can be 4h2π,5h2π,6h2π.
Hence as per the given options, D is the correct answer.