Question 705029

The eigenvalue of $\hat{L}^2$ is $l(l+1)\hbar^2$. For $l=1$, the eigenvalue is $1(1+1)\hbar^2 = 2\hbar^2$. The states corresponding to $l=1$ are $|\psi_{210}\rangle$ and $|\psi_{311}\rangle$. The probability is the sum of the squares of the coefficients: $(\sqrt{3/7})^2 + (\sqrt{4/21})^2 = 3/7 + 4/21 = 9/21 + 4/21 = 13/21$.