| A charged particle oscillates about its mean equilibrium position with a frequency of \[{{10}^{9}}\text{ }Hz\]. |
| The electromagnetic waves produced |
A) will have frequency of \[{{10}^{6}}\text{ }Hz\]
B) will have frequency of \[2\times {{10}^{3}}\ Hz\]
C) will have wavelength of 0.3 m
D) fall in the region of U.V. waves
Correct Answer: C
Solution :
[c] The frequency of the charged particles oscillates about its mean equilibrium position \[={{10}^{9}}Hz\]. Vibrating particle produces electric and magnetic field. So, frequency of electromagnetic waves produced by the charged particle is\[v={{10}^{9}}Hz\]. Wavelength \[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{{{10}^{9}}}=0.3m\] Since the range of radio waves is between 10 Hz to \[{{10}^{12}}Hz\] and hence \[{{10}^{9}}Hz\] lies in region of radio waves.
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