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Determine the frequency of revolution of the electron in $2^{nd}$ Bohr's orbit in hydrogen atom

$\begin{array}{1 1}(a)\;9.3\times 10^{15}Hz&(b)\;8.28\times 10^{16}Hz\\(c)\;7.32\times 10^{16}Hz&(d)\;8.13\times 10^{16}Hz\end{array}$

1 Answer

Frequency = $\large\frac{1}{\text{Time period}}$
Time period = $\large\frac{\text{Total distance covered}}{\text{Velocity}}= \large\frac{2\pi r}{v}$
Frequency = $\large\frac{v}{2\pi r}$
The velocity $(v_2)$ and radius $(r_2)$ of the $2^{nd}$ Bohr's orbit :
$r_n=(0.53\times 10^{-10}n^2/z)m$
$r_2=(0.53\times 10^{-10}(2^2)m = 2.12\times 10^{-100}m$
$v_n=2.165\times 10^6(z/n)m/s$
$v_2=2.165\times 10^6(1/2)=1.082\times 10^6m/s$
Frequency=$\Large\frac{v_2}{2\pi r_2}=\frac{1.082\times 10^6}{2(\pi)(2.12\times 10^{-10})}$
$v=8.13\times 10^{16}Hz$
Hence (d) is the correct option.
answered Jan 9, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1

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