The de-Broglie's wavelength of an electron in the $4^{\text {th }}$ orbit is ________ $\pi \mathrm{a}_0$. ($\mathrm{a}_0=$ Bohr's radius)

<p>The de-Broglie wavelength of an electron can be expressed through its relationship with the principal quantum number $n$ in a Bohr orbit. According to de Broglie, the wavelength $ \lambda $ of an electron moving in an orbit can be related to its momentum. However, when considering an electron in an atom, specifically within the Bohr model, we can relate the de-Broglie wavelength to the circumference of the orbit it moves in. The formula representing the de Broglie wavelength of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is:</p> <p>$\lambda_n = \frac{2 \pi r_n}{n}$</p> <p>where</p> <ul> <li>$ \lambda_n $ is the de-Broglie wavelength of the electron in the $n^{\text{th}}$ orbit,</li> <li>$ r_n $ is the radius of the $n^{\text{th}}$ orbit,</li> <li>$ n $ is the principal quantum number (which is 4 in this case).</li> </ul> <p>According to Bohr's theory, the radius of the $n^{\text{th}}$ orbit is given by:</p> <p>$r_n = n^2 a_0$</p> <p>where $ a_0 $ is the Bohr radius.</p> <p>For the $4^{\text{th}}$ orbit, $n = 4$, so the radius is:</p> <p>$r_4 = 4^2 a_0 = 16 a_0$</p> <p>Substituting $ r_4 $ into the de-Broglie wavelength formula, we get:</p> <p>$\lambda_4 = \frac{2 \pi (16 a_0)}{4} = 8 \pi a_0$</p>