To determine the value of \(x\) in the orbital angular momentum of an electron in the 3s orbital, we start by understanding the quantum numbers involved. The orbital angular momentum \(L\) is given by the formula: \(L=\sqrt{l(l+1)}\frac{h}{2\pi}\), where \(l\) is the azimuthal (angular momentum) quantum number. The problem specifies a 3s orbital. In quantum mechanics, for an s orbital, \(l=0\). Substituting \(l=0\) into the formula, we have: \(L=\sqrt{0(0+1)}\frac{h}{2\pi}=0\cdot\frac{h}{2\pi}=0\). Therefore, the orbital angular momentum of an electron in a 3s orbital is 0, implying \(x=0\). This value, \(x=0\), fits precisely within the given range of 0,0.
