If $a$₀ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ) of the electron present in the second orbit of hydrogen atom? [n : any integer]

<p>Bohr radius of hychogon atom $\rightarrow a_0$</p> <p>According to Bohr, the equation used to calculate the angular momentum of an election in a hydrogen atom is</p> <p>$\mathrm{mvr=\frac{nh}{2\pi}}$ ..... (1)</p> <p>$m \rightarrow$ mass of electron</p> <p>$v \rightarrow$ velocity of electron</p> <p>$r \rightarrow$ radius of the orbit</p> <p>$n \rightarrow$ orbit. number. in. which electron is present.</p> <p>Given that; election is present in second orbit, $n=2$</p> <p>The radius of the second orbit $r_2=a_0\times2^2=4a_0$</p> <p>General formula for radius of $n^{th}$ orbit,</p> <p>$r_n=a_0\times n^2$</p> <p>From (1)</p> <p>$$\begin{aligned} & m v r=n \frac{h}{2 \pi} \\ & 2 \pi r=n \frac{h}{m v} \end{aligned}$$</p> <p>$\frac{h}{m v}=\lambda$ (de Broglie relation ship, $\lambda \rightarrow$ de Broglie Wavelength</p> <p>So, $2 \pi r=n \lambda$</p> <p>For the electron in the second orbit, $2 \pi r_2=n \lambda$</p> <p>Substitute for $r_2$,</p> <p>$$\begin{aligned} & 2 \pi \times 4 a_0=n \lambda \\ & 8 \pi a_0=n \lambda \\ & \therefore \lambda=\frac{8 \pi a_0}{n} \end{aligned}$$</p> <p>Correct answer: Option 1) $\frac{8 \pi a_0}{n}$</p>