De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

To find the de-Broglie wavelength of an electron orbiting in the $n=2$ state of a hydrogen atom, we can use the relation between the orbital perimeter and the de-Broglie wavelength.

According to Bohr's quantization condition, the angular momentum of an electron in the $n$-th orbit is given by:
$mvr=\frac{nh}{2\pi }$
where:
$m$ is the mass of the electron,
$v$ is the velocity of the electron in the $n$-th orbit,
$r$ is the radius of the orbit,
$n$ is the principal quantum number, and
$h$ is Planck's constant.

We can rewrite the above equation as:
$2\pi r=n\left(\frac{h}{mv}\right)$

By de-Broglie's hypothesis, the wavelength $\lambda$ associated with the moving electron is:
$\lambda =\frac{h}{mv}$

Substituting $\lambda$ into the quantization equation, we get:
$2\pi r=n\lambda$

Therefore, the de-Broglie wavelength $\lambda$ is given by:
$\lambda =\frac{2\pi r}{n}$

For the hydrogen atom, the radius of the $n$-th orbit is given by:
$r=r_0{n}^2$
where $r_0$ is the Bohr radius, given as 0.0529 nm (or approximately 0.052 nm in the question).

For the $n=2$ state:
$r=0.052\text{nm}\times 2^2=0.052\text{nm}\times 4=0.208\text{nm}$

Now, we substitute the value of $r$ and $n=2$ back into the expression for $\lambda$:
$\lambda =\frac{2\times \pi \times 0.208\text{nm}}{2}$
$\lambda =\pi \times 0.208\text{nm}$
$\lambda \approx 3.1416\times 0.208\text{nm}\approx 0.653\text{nm}$

Using $r_0=0.0529\text{nm}$ provides a more precise result:
$r=0.0529\text{nm}\times 4=0.2116\text{nm}$
$\lambda =\pi \times 0.2116\text{nm}\approx 3.1416\times 0.2116\text{nm}\approx 0.665\text{nm}\approx 0.67\text{nm}$

Thus, the de-Broglie wavelength of the electron in the $n=2$ orbit is closest to 0.67 nm.