If $\mathrm{a}_0$ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength $(\lambda)$ of the electron present in the second orbit of hydrogen atom? [n : any integer]

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Aicharya · Accepted Answer

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

If \(\mathrm{a}_0\) is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength…

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