The de-Broglie’s wavelength of an electron in the 4th orbit is _______ πa₀. (a₀ = Bohr’s radius)

The correct answer is 8.

To find the de-Broglie wavelength of an electron in the 4th orbit of a hydrogen atom, we can use Bohr's quantization condition and the de-Broglie hypothesis.

According to Bohr's quantization of angular momentum, the angular momentum of an electron in the n-th orbit is given by:

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

where:
m is the mass of the electron,
v is the velocity of the electron in the n-th orbit,
r_n is the radius of the n-th orbit,
h is Planck's constant, and
n is the principal quantum number.

Rearranging the equation to find momentum (p = mv):

$p=mv=\frac{nh}{2\pi r_n}$

According to de-Broglie's relation, the wavelength (λ) of the electron is:

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

Substitute the expression for momentum (p) into the de-Broglie wavelength equation:

$\lambda =\frac{h}{\left(\frac{nh}{2\pi r_n}\right)}=\frac{2\pi r_n}{n}$

This gives the relation between the orbit's circumference and the de-Broglie wavelength:
$n\lambda =2\pi r_n$

For a hydrogen atom, the radius of the n-th orbit is given by:
$r_n=n^2{a}_0$
where a₀ is the Bohr radius.

Substituting this value of r_n into our equation for λ:

$n\lambda =2\pi \left(n^2{a}_0\right)$

Simplifying for λ:

$\lambda =2\pi n{a}_0$

For the 4th orbit, we set n = 4:

$\lambda =2\pi \left(4\right)a_0=8\pi a_0$

Therefore, the de-Broglie wavelength of the electron in the 4th orbit is 8 πa₀.