We write $ x^2 + y^2 + z^2 = 1 $ as $ x^2 + ay^2 + (1 - a) y^2 + z^2 = 1, $ where $ a $ is some real number, $ 0 \le a \le 1, $ that is to be determined. Then by AM-GM, \[1 = x^2 + ay^2 + (1 - a) y^2 + z^2 \ge 2xy \sqrt{a} + 2yz \sqrt{1 - a}.\]To make this look more like $ \lambda xy + yz, $ we choose $ a $ so that the coefficients are proportional, i.e. \[\frac{2 \sqrt{a}}{\lambda} = 2 \sqrt{1 - a}.\]Then $ \sqrt{a} = \lambda \sqrt{1 - a}, $ so $ a = \lambda^2 (1 - a) $. Solving for $ a, $ we find $ a = \frac{\lambda^2}{\lambda^2 + 1} $. Therefore, \[\frac{2 \lambda xy}{\sqrt{\lambda^2 + 1}} + \frac{2yz}{\sqrt{\lambda^2 + 1}} \le 1,\]so \[\lambda xy + yz \le \frac{\sqrt{\lambda^2 + 1}}{2}.\]We want this to equal $ \frac{\sqrt{5}}{2}, $ so $ \lambda = \boxed{2} $. Equality occurs when $ x = \frac{\sqrt{10}}{5}, $ $ y = \frac{\sqrt{2}}{2}, $ and $ z = \frac{\sqrt{10}}{10} $.