If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 a0 (a0 is Bohr…

🧠 The Orbit Circumference According to Bohr and de Broglie, for an orbit to be stable, the path taken ($2\pi r$) must be an exact multiple of the electron's wavelength.

🗺️ The Geometric Equation

1. The Core Law: $n\lambda = 2\pi r_n$.
2. The Bohr Radius: $r_n = a_0 \frac{n^2}{Z}$.
3. Plugging into the Wave Condition: $n (1.5\pi a_0) = 2\pi \left( a_0 \frac{n^2}{Z} \right)$
4. Simplifying: Cancel $\pi, a_0,$ and one $n$: $1.5 = 2 \frac{n}{Z}$ $\frac{n}{Z} = \frac{1.5}{2} = 0.75$

⚡ The $1.5/2$ Quick Fix The ratio $n/Z$ is essentially a measurement of how "stretched" the wavelength is relative to the radius. Since $2\pi r$ is $2\times$ the radius factor, the result is simply half of the $1.5$ coefficient.

⚠️ Common Traps Forgetting that $r_n$ scales with $n^2$. If you use $r_n = a_0 n/Z$, you will get an incorrect value of $1.5 cdot 2$.

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