If the radius of the first Bohr orbit is ' $r$ ' then the de-Broglie wavelength of the electron in the…
If the radius of the first Bohr orbit is ' $r$ ' then the de-Broglie wavelength of the electron in the $4^{\text {th }}$ orbit will be
- $4 \pi r$
- $6 \pi r$
- $8 \pi r$
- $\frac{\pi \mathrm{r}}{4}$
Solution
According to Bohr's second postulate, $\frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}}$
$\therefore \quad$ de-Broglie wavelength, $\lambda_{\mathrm{n}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}_{\mathrm{n}}}{\mathrm{n}}$
Also, $r_n \propto n^2$
$\therefore \quad$ The de-Broglie wavelength of the electron in the $4^{\text {th }}$ orbit is:
$\begin{aligned}
\quad \lambda_4 & =\frac{2 \pi r_4}{4}=\frac{2 \pi \times(16 r)}{4} \\
\therefore \quad \lambda_4 & =8 \pi \mathrm{r}
\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
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