1. Stating the problem: A binary operation $*$ is defined on the set $\mathbb{R}$ of real numbers by $a * b = \frac{ab}{4}$. Find $\sqrt{2} * \sqrt{6}$.\n2. Formula and rules: Use the operation definition $a*b=\frac{ab}{4}$ and the property $\sqrt{u}\,\sqrt{v}=\sqrt{uv}$ for nonnegative $u,v$.\n3. Compute using the definition: $\sqrt{2} * \sqrt{6}=\frac{\sqrt{2}\cdot\sqrt{6}}{4}$.\n4. Multiply the square roots: $\sqrt{2}\cdot\sqrt{6}=\sqrt{12}$.\n5. Simplify the radical by factoring: $\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}$.\n6. Substitute back into the fraction: $\frac{\sqrt{12}}{4}=\frac{2\sqrt{3}}{4}$.\n7. Show cancellation of common factors and simplify: $$\frac{2\sqrt{3}}{4}=\frac{\cancel{2}\sqrt{3}}{\cancel{4}}=\frac{\sqrt{3}}{2}$$\n8. Final answer: $\sqrt{2} * \sqrt{6}=\frac{\sqrt{3}}{2}$, which matches option C.\n