Radius of first orbit of hydrogen atom is ‘4x’ A°. Wavelength of electron revolving in fourth orbit of hydrogen atom is

# Radius of first orbit of hydrogen atom is '4x' A°. Wavelength of electron revolving in fourth orbit of hydrogen atom is

1. A

$16{\mathrm{\pi xA}}^{\mathrm{o}}$

2. B

$32{\mathrm{\pi xA}}^{\mathrm{o}}$

3. C

$64{\mathrm{\pi xA}}^{\mathrm{o}}$

4. D

$8{\mathrm{\pi xA}}^{\mathrm{o}}$

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### Solution:

$\begin{array}{l}2\mathrm{\pi r}=\mathrm{n\lambda }\\ 2\mathrm{\pi }\left({\mathrm{r}}_{\mathrm{o}}{\mathrm{n}}^{2}{\mathrm{A}}^{\mathrm{o}}\right)=\mathrm{n\lambda }\\ 2\mathrm{\pi }\left(4\mathrm{x}×4{\mathrm{A}}^{\mathrm{o}}\right)=\mathrm{\lambda }\\ 32{\mathrm{\pi xA}}^{\mathrm{o}}=\mathrm{\lambda }\end{array}$

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