If the radius of the first orbit of hydrogen atom is a0, then de Broglie's wavelength of electron in the 3rd orbit is:

🧠 The Circumference Constraint For an electron wave to be stable in a Bohr orbit, the orbit's path must perfectly accommodate a whole number of wavelengths ($n\lambda = 2\pi r$).

🗺️ The Wave Fit Derivation

1. Radius Scaling: $r_n = a_0 \cdot n^2$. For the $3^{\text{rd}}$ orbit ($n=3$), $r_3 = 9a_0$.
2. Quantization Rule: $2\pi r_3 = 3\lambda$.
3. Calculation: $2\pi (9a_0) = 3\lambda \implies 18\pi a_0 = 3\lambda$. $\lambda = 6\pi a_0$

⚡ The $2n\pi a_0$ Trick Since $2\pi(n^2 a_0) = n\lambda$, the wavelength is always $2\pi n a_0$. For $n=3$, it's $6\pi a_0$. For $n=4$, it's $8\pi a_0$.

⚠️ Common Traps Don't use $\lambda = h/mv$ unless you want a headache with constants. The geometric "loop" method is much faster.

$\boxed{\text{Answer: (C)}}$