The de Broglie wavelength of an electron in the 4th Bohr orbit is:

🧠 The Standing Wave Condition For an electron to exist in a stable Bohr orbit, the circumference of that orbit must precisely fit an integral number of de Broglie wavelengths ($n\lambda = 2\pi r$). If it doesn't fit, the wave destroys itself through interference.

🗺️ The Wave Fit Path

1. The Quantization Rule: $n\lambda = 2\pi r_n$ For the $4^{\text{th}}$ orbit ($n=4$): $4\lambda = 2\pi r_4 \implies \lambda = \frac{\pi r_4}{2}$
2. The Radius Scaling: In Hydrogen, $r_n = a_0 n^2$. For $n=4$: $r_4 = a_0 (4)^2 = 16 a_0$
3. The Final Substitution: $\lambda = \frac{\pi (16 a_0)}{2} = 8\pi a_0$

⚡ The $2n\pi a_0$ Shortcut Remember that since $2\pi (a_0 n^2) = n\lambda$, the wavelength is simply $2\pi n a_0$. For $n=4$, it's $8\pi a_0$. For $n=3$, it's $6\pi a_0$.

⚠️ Common Traps Don't use the simple $n\lambda$ product without accounting for the $n^2$ growth of the radius. The orbit gets much bigger as $n$ increases!

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