In the Bohr model of the hydrogen atom, the quantization condition for the angular momentum states that the angular momentum is an integer multiple of $$ \hbar $$ (reduced Planck's constant). Mathematically, this is given by:

$$ mvr = n\hbar $$

where:

$ m $ is the electron mass,

$ v $ is the electron velocity,

$ r $ is the radius of the orbit,

$ n $ is the principal quantum number,

$ \hbar = \frac{h}{2\pi} $ (where $ h $ is Planck's constant).

According to de-Broglie, an electron in a stable orbit behaves like a standing wave. The condition for standing waves is that the circumference of the orbit must be an integer multiple of the electron's de-Broglie wavelength $ \lambda $, which is given by:

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

For the wave to be stable, the circumference of the orbit $ 2\pi r $ must equal $ n $ times the de-Broglie wavelength:

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

Substituting for $ \lambda $ gives:

$$ 2\pi r = n \left(\frac{h}{mv}\right) $$

This equation is consistent with the quantization condition. It shows that the number of de-Broglie wavelengths that can fit in the orbit is equal to the principal quantum number $ n $.

For the second Bohr orbit, $ n = 2 $. Hence, the number of de-Broglie wavelengths contained in the second Bohr orbit is:

Option B: 2