"If the radius of the first orbit of hydrogen atom is $\alpha_{0}$, then de Broglie's wavelength of electron in $3^{\text {rd }}$ orbit is :"

Question

Accepted Answer

"

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:

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

"

If the radius of the first orbit of hydrogen atom is $\alpha_{0}$, then de Broglie's wavelength of electron in $3^{\text {rd }}$ orbit is :