In Bohr's model, the angular momentum of an electron in an orbit is quantized, which can be represented as:

$$mvr = n \cdot \frac{h}{2\pi}$$

Where:

• (m) is the mass of the electron,
• (v) is the velocity of the electron,
• (r) is the radius of the nth orbit, and
• (n) is the principal quantum number ((n = 1) for the first orbit, (n = 2) for the second orbit, and so on).

The de Broglie wavelength ($\lambda$) is given by:

$$\lambda = \frac{h}{mv}$$

For the nth orbit, the radius (r) is $n^2$ times the radius of the first orbit ($r_0$, or $\alpha_0$ in your notation):

$$r = n^2 \cdot \alpha_0$$

The circumference of the nth orbit is $2\pi r$, which can be written as:

$$2\pi r = n \cdot \lambda$$

Substituting the expression for (r):

$$2\pi \cdot n^2 \cdot \alpha_0 = n \cdot \lambda$$

Solving for $\lambda$ gives:

$$\lambda = 2\pi \cdot n \cdot \alpha_0$$

For the 3rd orbit ((n = 3)), this simplifies to:

$$\lambda = 6\pi \cdot \alpha_0$$