If a0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength () of the electron present…

🧠 The Circumference Quantization Louis de Broglie justified Bohr's "magic" orbits by suggesting that electrons form standing waves. For a wave to "survive" in a circle, the total path length (circumference) must fit exactly an integral number of wavelengths. $2\pi r = n\lambda$

🗺️ Building the Wavelength

1. Orbit ($n$): We are in the $n=2$ orbit.
2. Radius ($r$): From Bohr theory, $r_n = a_0 n^2$. For $n=2$: $r_2 = a_0 (2)^2 = 4a_0$.
3. Equating for $\lambda$: $2\pi r_2 = 2 \lambda$ $2\pi (4a_0) = 2\lambda \implies 8\pi a_0 = 2\lambda$ $\lambda = 4\pi a_0$
4. Matching the Formula: The options use the general form $\lambda = \frac{8\pi a_0}{n}$. If we substitute $n=2$ here, we get $\frac{8\pi a_0}{2} = 4\pi a_0$.

⚡ The $2\pi a_0 n$ Shortcut For Hydrogen, the de-Broglie wavelength in any orbit $n$ is simply $\lambda_n = 2\pi a_0 n$. For $n=2$, $\lambda = 4\pi a_0$. Just find the option that simplifies to this!

⚠️ Radius vs. Circumference Don't say the wavelength is the radius. The wavelength is part of the length of the circle ($2\pi r$). For the second orbit, two full wave-cycles must wrap around the nucleus.

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