The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is:

We begin with the de-Broglie relation, which states that the wavelength associated with a moving particle is given by

$$\lambda=\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p=mv$$ is the linear momentum of the particle of mass $$m$$ moving with speed $$v$$.

For an electron revolving around the nucleus in a Bohr orbit we additionally have Bohr’s angular-momentum quantisation condition, stated as

$$mvr=\frac{nh}{2\pi},$$

where $$r$$ is the radius of the orbit and $$n$$ is the principal quantum number (an integer).

Rearranging this Bohr condition to express the momentum $$mv$$ gives

$$mv=\frac{nh}{2\pi r}.$$

We now substitute this value of $$mv$$ into the de-Broglie formula. Doing every algebraic step explicitly we get

$$\lambda=\frac{h}{mv} =\frac{h}{\displaystyle\frac{nh}{2\pi r}} =h \times \frac{2\pi r}{nh} =\frac{2\pi r}{n}.$$

We are interested in the first Bohr orbit, so we put $$n=1$$. This immediately yields

$$\lambda=\frac{2\pi r}{1}=2\pi r_1,$$

where $$r_1$$ is the radius of the first Bohr orbit.

The well-known expression for the Bohr radius is

$$r_1=a_0=0.529\ \text{Å},$$

with $$1\ \text{Å}=10^{-10}\ \text{m}$$. No unit conversion is necessary because the options are already expressed in ångströms.

Substituting $$r_1=0.529\ \text{Å}$$ into the wavelength expression we obtain

$$\lambda = 2\pi \times 0.529\ \text{Å}.$$

This result matches exactly with Option B.

Hence, the correct answer is Option 2.