Question:
The radius of second orbit of hydrogen atom is 2.12 Å and the velocity of the electron revolving in this orbit is $1.1\times10^6 ms^{-1}$. According to classical electromagnetic theory, the initial frequency of light emitted by this electron is nearly}
The radius of second orbit of hydrogen atom is 2.12 Å and the velocity of the electron revolving in this orbit is $1.1\times10^6 ms^{-1}$. According to classical electromagnetic theory, the initial frequency of light emitted by this electron is nearly}
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In classical atomic theory, the emitted radiation frequency equals the orbital frequency of the revolving electron.
Updated On: Jun 17, 2026
- $2.05\times10^{14}Hz$
- $16.50\times10^{14}Hz$
- $4.15\times10^{14}Hz$
- $8.25\times10^{14}Hz$
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The Correct Option is D
Solution and Explanation
Concept:
According to classical theory, an accelerating charge emits electromagnetic radiation having frequency equal to the frequency of revolution.
Therefore,
\[
f=\frac{v}{2\pi r}
\]
Step 1: Convert radius into SI unit.
\[ r=2.12\AA \] \[ r=2.12\times10^{-10}m \]
Step 2: Calculate orbital frequency.
\[ f=\frac{1.1\times10^6} {2\pi(2.12\times10^{-10})} \] \[ f\approx8.25\times10^{14}Hz \] \[ \boxed{8.25\times10^{14}Hz} \]
Step 1: Convert radius into SI unit.
\[ r=2.12\AA \] \[ r=2.12\times10^{-10}m \]
Step 2: Calculate orbital frequency.
\[ f=\frac{1.1\times10^6} {2\pi(2.12\times10^{-10})} \] \[ f\approx8.25\times10^{14}Hz \] \[ \boxed{8.25\times10^{14}Hz} \]
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